Chapter 4 part 1 WORK AND ENERGY S.Chand Solution Class 9

 Very Short Answer Type Questions

1. How much work is done when a body of mass m is raised to a height h above the ground ?

The work done when a body of mass $�$ is raised to a height $ℎ$ above the ground is given by the product of the force applied and the displacement, which in this case is the height lifted: $\text{Work}=\text{Force}\times \text{Displacement}$. Since the force required to lift an object is equal to its weight $��$, where $�$ is the acceleration due to gravity, the work done is $\text{Work}=��ℎ$.

2. State the SI unit of work.

The SI unit of work is the joule (J).

3. Is work a scalar or a vector quantity ?

Work is a scalar quantity.

4. Define 1 joule of work.

One joule of work is defined as the work done when a force of one newton is applied over a distance of one meter in the direction of the force.

5. What is the condition for a force to do work on a body ?

The condition for a force to do work on a body is that there must be displacement of the point of application of the force in the direction of the force.

6. Is energy a vector quantity ?

No, energy is a scalar quantity.

7. What are the units of (a) work, and (b) energy ?

(a) The units of work are joules (J). (b) The units of energy are also joules (J).

8. What is the work done against gravity when a body is moved horizontally along a frictionless surface ?

When a body is moved horizontally along a frictionless surface, no work is done against gravity because the force required to overcome gravity is perpendicular to the displacement.

9. By how much will the kinetic energy of a body increase if its speed is doubled ?

If the speed of a body is doubled, its kinetic energy increases by a factor of four.

10. Write an expression for the kinetic energy of a body of mass m moving with a velocity v.

The kinetic energy $��$ of a body of mass $�$ moving with a velocity $�$ is given by the expression: $��=\frac{1}{2}�{�}^2$.

11. If the speed of a body is halved, what will be the change in its kinetic energy ?

If the speed of a body is halved, its kinetic energy decreases to one-fourth of its original value.

12. On what factors does the kinetic energy of a body depend ?

The kinetic energy of a body depends on its mass and the square of its velocity.

13. Which would have a greater effect on the kinetic energy of an object : doubling the mass or doubling the

velocity ?

Doubling the velocity would have a greater effect on the kinetic energy of an object compared to doubling the mass. This is because kinetic energy is proportional to the square of velocity, whereas it is directly proportional to mass.

14. How fast should a man of 50 kg run so that his kinetic energy be 625 J ?

1. To find the velocity $�$ required for a kinetic energy of 625 J for a man of 50 kg, we use the kinetic energy formula: $��=\frac{1}{2}�{�}^2$. $625=\frac{1}{2}\times 50\times {�}^2$ ${�}^2=\frac{2\times 625}{50}=25$ $�=\sqrt{25}=5\text{m/s}$

15. State whether the following objects possess kinetic energy, potential energy, or both : (a) A man climbing a hill (b) A flying aeroplane (c) A bird running on the ground (d) A ceiling fan in the off position (e) A stretched spring lying on the ground.

(a) A man climbing a hill possesses both kinetic energy (due to motion) and potential energy (due to height). (b) A flying airplane possesses kinetic energy. (c) A bird running on the ground possesses kinetic energy. (d) A ceiling fan in the off position possesses potential energy. (e) A stretched spring lying on the ground possesses potential energy. 16. Two bodies A and B of equal masses are kept at heights of h and 2h respectively. What will be the ratio of their potential energies ?

The potential energy of an object at a height $ℎ$ is given by $��ℎ$, where $�$ is the mass, $�$ is the acceleration due to gravity, and $ℎ$ is the height. Therefore, for body A, the potential energy is $��ℎ$, and for body B, it is $�(2ℎ)�=2��ℎ$. So, the ratio of their potential energies is: $\frac{\text{Potential energy of B}}{\text{Potential energy of A}}=\frac{2��ℎ}{��ℎ}=2$ 17. What is the kinetic energy of a body of mass 1 kg moving with a speed of 2 m/s ?

The kinetic energy $��$ of a body of mass $�$ moving with a speed $�$ is given by the formula $��=\frac{1}{2}�{�}^2$. Substituting the given values, we get: $��=\frac{1}{2}\times 1\times (2{)}^2=2\text{J}$ 18. Is potential energy a vector or a scalar quantity ?

Potential energy is a scalar quantity. 19. A load of 100 kg is pulled up by 5 m. Calculate the work done. (g = 9.8 m/s2)

The work done when pulling the load up by 5 m is given by the formula $\text{Work}=\text{Force}\times \text{Distance}$. The force required to overcome gravity is $�=��$, where $�$ is the mass and $�$ is the acceleration due to gravity. Substituting the given values, we get: $\text{Work}=�\times �=��ℎ=100\times 9.8\times 5=4900\text{J}$ 20. State whether the following statement is true or false : The potential energy of a body of mass 1 kg kept at a height of 1 m is 1 J.

False. The potential energy of a body of mass 1 kg kept at a height of 1 m is $��ℎ=1\times 9.8\times 1=9.8\text{J}$. 21. What happens to the potential energy of a body when its height is doubled ?

When the height of a body is doubled, its potential energy also doubles, assuming the mass and gravitational acceleration remain constant. 22. What kind of energy is possessed by the following ? (a) A stone kept on roof-top. (b) A running car. (c) Water stored in the reservoir of a dam. (d) A compressed spring. (e) A stretched rubber band.

(a) A stone kept on a rooftop possesses gravitational potential energy. (b) A running car possesses kinetic energy and chemical potential energy (in its fuel). (c) Water stored in the reservoir of a dam possesses gravitational potential energy. (d) A compressed spring possesses elastic potential energy. (e) A stretched rubber band possesses elastic potential energy.

23. Fill in the following blanks with suitable words :

(a) Work is measured as a product of force and displacement. (b) The work done on a body moving in a circular path is zero. (c) 1 joule is the work done when a force of one newton moves an object through a distance of one meter in the direction of the force. (d) The ability of a body to do work is called energy. The ability of a body to do work because of its motion is called kinetic energy. (e) The sum of the potential and kinetic energies of a body is called mechanical energy.

Short Answer Type Questions

24. What are the quantities on which the amount of work done depends ? How are they related to work ?

The amount of work done depends on two main quantities:

1. Force: The force applied to an object determines the amount of work done. Work is directly proportional to the magnitude of the force. The greater the force applied, the more work is done.

2. Displacement: The displacement of the object in the direction of the force is also crucial in determining the amount of work done. Work is directly proportional to the magnitude of displacement. The greater the distance over which the force is applied, the more work is done.

The relationship between work ($�$), force ($�$), and displacement ($�$) is given by the formula:

$�=�\times �$

This formula states that work is equal to the product of force and displacement, indicating that both force and displacement play a significant role in determining the amount of work done.

25. Is it possible that a force is acting on a body but still the work done is zero ? Explain giving one example.

Yes, it is indeed possible for a force to act on a body, yet the work done is zero. This occurs when the force and the displacement of the body are perpendicular to each other. In such cases, the force may cause a change in the direction of motion or the shape of the object, but it doesn't result in a displacement in the direction of the force, and hence, no work is done.

Example: Consider a person holding a heavy box and walking horizontally along a straight path. The gravitational force is acting vertically downward on the box, but the displacement of the box is horizontal. Since the force and the displacement are perpendicular to each other, the work done by the gravitational force is zero. The person is exerting force to counteract gravity, preventing the box from falling, but the work done by the gravitational force is zero because there's no displacement in the vertical direction.

26. A boy throws a rubber ball vertically upwards. What type of work, positive or negative, is done :

(a) by the force applied by the boy ?

(b) by the gravitational force of earth ?

Let's analyze the situation:

(a) By the force applied by the boy:

- When the boy throws the rubber ball vertically upwards, he is applying a force on the ball in the upward direction.

- Since the displacement of the ball is also in the upward direction, the angle between the force and the displacement is \( 0^\circ \).

- Work done by a force is positive when the force and displacement are in the same direction.

- Therefore, the work done by the force applied by the boy is positive.

(b) By the gravitational force of Earth:

- As the rubber ball moves upwards, the gravitational force of the Earth acts downwards.

- The displacement of the ball is upwards, opposite to the direction of the gravitational force.

- Work done by a force is negative when the force and displacement are in opposite directions.

- Therefore, the work done by the gravitational force of Earth is negative.

In summary:

(a) The work done by the force applied by the boy is positive.

(b) The work done by the gravitational force of Earth is negative.

27. Write the formula for work done on a body when the body moves at an angle to the direction of force. Give

the meaning of each symbol used.

When the body moves at an angle to the direction of the force, the formula for work done ($�$) is given by:

$�=�\times �\times \cos (�)$

Where:

• $�$ is the work done on the body (in joules, J).
• $�$ is the magnitude of the force applied on the body (in newtons, N).
• $�$ is the displacement of the body (in meters, m).
• $�$ is the angle between the force vector and the displacement vector (in degrees). $�$ is measured from the direction of the force to the direction of displacement.

This formula incorporates the angle between the force and displacement vectors through the cosine of the angle $�$. When $�=0^{\circ }$, indicating that the force and displacement are in the same direction, $\cos (0^{\circ })=1$, and the formula reduces to $�=�\times �$, which represents the case of work done when the force and displacement are parallel.

This formula accounts for the fact that only the component of the force in the direction of the displacement contributes to the work done on the body. The cosine of the angle $�$ helps in finding this component, ensuring that the work done is correctly calculated when the force and displacement are not aligned.

28. How does the kinetic energy of a moving body depend on its (i) speed, and (ii) mass ?

The kinetic energy ($��$) of a moving body depends on both its speed and mass.

(i) Speed:

• Kinetic energy is directly proportional to the square of the speed ($�$) of the body. Mathematically, $��$ is given by the formula: $��=\frac{1}{2}�{�}^2$ Where:
  • $�$ is the mass of the body.
  • $�$ is the speed of the body.
• This means that if the speed of the body doubles, its kinetic energy increases by a factor of four. Similarly, if the speed triples, the kinetic energy increases by a factor of nine.

(ii) Mass:

• Kinetic energy is directly proportional to the mass ($�$) of the body.
• This relationship is linear; doubling the mass of the body doubles its kinetic energy, assuming the speed remains constant.
• The kinetic energy of a body is directly proportional to its mass. So, if two bodies have the same speed but different masses, the one with a greater mass will possess more kinetic energy.

In summary, the kinetic energy of a moving body depends quadratically on its speed and linearly on its mass.

29. Give one example each in which a force does (a) positive work (b) negative work, and (c) zero work.

Certainly, here are examples for each scenario:

(a) Positive Work:

• Example: A person pushes a box across a floor in the direction of motion.
• Explanation: When the person applies a force on the box in the direction of motion, the displacement of the box is also in the same direction as the force. As a result, the work done by the person's force is positive.

(b) Negative Work:

• Example: Frictional force acting on a sliding block.
• Explanation: When a block slides across a rough surface, the frictional force opposes the direction of motion. Therefore, the displacement of the block is in the direction opposite to the frictional force. As a result, the work done by the frictional force is negative.

(c) Zero Work:

• Example: Carrying a backpack while walking in a straight line.
• Explanation: When a person carries a backpack while walking in a straight line at a constant speed, the force exerted by the person is in the vertical direction (against gravity), but the displacement of the backpack is horizontal. Since the force and displacement are perpendicular to each other, the work done by the gravitational force is zero.

30. A ball of mass 200 g falls from a height of 5 metres. What is its kinetic energy when it just reaches the

ground ? (g = 9.8 m/s2).

To find the kinetic energy ($��$) of the ball just as it reaches the ground, we can use the formula for gravitational potential energy ($��$) and the principle of conservation of energy.

Given:

• Mass of the ball, $�=200$ g $=0.2$ kg
• Height from which the ball falls, $ℎ=5$ m
• Acceleration due to gravity, $�=9.8$ m/s²

The gravitational potential energy ($��$) of the ball at height $ℎ$ is given by:

$��=��ℎ$

Substituting the given values:

$��=0.2\times 9.8\times 5$

$��=9.8\text{J}$

At the moment just before it hits the ground, all of the potential energy ($��$) is converted into kinetic energy ($��$) due to the law of conservation of energy. Therefore:

$��=��=9.8\text{J}$

So, the kinetic energy of the ball just as it reaches the ground is $9.8\text{J}$.

31. Find the momentum of a body of mass 100 g having a kinetic energy of 20 J.

To find the momentum ($�$) of a body, we can use the formula relating kinetic energy ($��$) and momentum ($�$):

$��=\frac{1}{2}�{�}^2$

Where:

• $�$ is the mass of the body (in kg).
• $�$ is the velocity of the body (in m/s).

First, let's find the velocity ($�$) of the body using the given kinetic energy ($��$).

Given:

• Mass of the body, $�=100$ g $=0.1$ kg
• Kinetic energy, $��=20$ J

Using the formula $��=\frac{1}{2}�{�}^2$, we rearrange it to solve for $�$:

$�=\sqrt{\frac{2��}{�}}$

Substituting the given values:

$�=\sqrt{\frac{2\times 20}{0.1}}$

$�=\sqrt{400}$

$�=20\text{m/s}$

Now that we have the velocity ($�$), we can find the momentum ($�$) using the formula:

$�=��$

Substituting the mass ($�$) and velocity ($�$):

$�=0.1\times 20$

$�=2\text{kg}\cdot \text{m/s}$

So, the momentum of the body is $2\text{kg}\cdot \text{m/s}$.

32. Two objects having equal masses are moving with uniform velocities of 2 m/s and 6 m/s respectively.

Calculate the ratio of their kinetic energies.

The kinetic energy ($��$) of an object is given by the formula:

$��=\frac{1}{2}�{�}^2$

Where:

• $�$ is the mass of the object.
• $�$ is the velocity of the object.

Given:

• Two objects have equal masses, so ${�}_1={�}_2=�$.
• The velocities of the two objects are ${�}_1=2$ m/s and ${�}_2=6$ m/s.

Let's denote the kinetic energies of the two objects as $�{�}_1$ and $�{�}_2$.

For the first object: $�{�}_1=\frac{1}{2}�{�}_1^2=\frac{1}{2}�(2{)}^2=2�$

For the second object: $�{�}_2=\frac{1}{2}�{�}_2^2=\frac{1}{2}�(6{)}^2=18�$

Now, let's find the ratio of their kinetic energies:

$\frac{�{�}_2}{�{�}_1}=\frac{18�}{2�}=9$

So, the ratio of their kinetic energies is $9:1$.

33. A body of 2 kg falls from rest. What will be its kinetic energy during the fall at the end of

2 s ? (Assume g = 10 m/s2)

To find the kinetic energy ($��$) of the falling body at the end of 2 seconds, we can use the formula for kinetic energy:

$��=\frac{1}{2}�{�}^2$

Where:

• $�$ is the mass of the body (in kg).
• $�$ is the velocity of the body (in m/s).

Given:

• Mass of the body, $�=2$ kg
• Time taken, $�=2$ s
• Acceleration due to gravity, $�=10$ m/s² (assuming downward direction as positive)

We know that the velocity ($�$) of the body after falling for $�$ seconds under gravity can be calculated using the equation of motion:

$�=�+��$

Where:

• $�$ is the initial velocity (in this case, the body starts from rest, so $�=0$).
• $�$ is the acceleration due to gravity.
• $�$ is the time taken.

Substituting the given values:

$�=0+(10)(2)$ $�=20\text{m/s}$

Now, we can find the kinetic energy ($��$) using the formula:

$��=\frac{1}{2}�{�}^2$ $��=\frac{1}{2}(2)(20{)}^2$ $��=\frac{1}{2}\times 2\times 400$ $��=400\text{J}$

So, the kinetic energy of the body at the end of 2 seconds of fall is $400\text{J}$.

34. On a level road, a scooterist applies brakes to slow down from a speed of 10 m/s to 5 m/s. If the mass of the

scooterist and the scooter be 150 kg, calculate the work done by the brakes. (Neglect air resistance and

friction)

To calculate the work done by the brakes, we can use the work-energy principle. The work done by the brakes is equal to the change in kinetic energy of the scooter and the scooterist system.

Given:

• Initial velocity (${�}_{�}$) = 10 m/s
• Final velocity (${�}_{�}$) = 5 m/s
• Mass of the scooterist and scooter ($�$) = 150 kg

First, let's calculate the initial kinetic energy ($�{�}_{�}$) and final kinetic energy ($�{�}_{�}$):

$�{�}_{�}=\frac{1}{2}�{�}_{�}^2$ $�{�}_{�}=\frac{1}{2}\times 150\times (10{)}^2$ $�{�}_{�}=\frac{1}{2}\times 150\times 100$ $�{�}_{�}=7500\text{J}$

$�{�}_{�}=\frac{1}{2}�{�}_{�}^2$ $�{�}_{�}=\frac{1}{2}\times 150\times (5{)}^2$ $�{�}_{�}=\frac{1}{2}\times 150\times 25$ $�{�}_{�}=1875\text{J}$

Now, the work done by the brakes ($�$) is the change in kinetic energy:

$�=�{�}_{�}-�{�}_{�}$ $�=1875-7500$ $�=-5625\text{J}$

The negative sign indicates that work is done by an external force (the brakes) to decrease the kinetic energy of the system. In other words, the brakes exert a force in the direction opposite to the motion of the scooter, which is why the work done is negative.

35. A man drops a 10 kg rock from the top of a 5 m ladder. What is its speed just before it hits the ground ?

What is its kinetic energy when it reaches the ground ? ( g = 10 m/s2)

To find the speed of the rock just before it hits the ground, we can use the equation of motion under gravity:

${�}^2={�}^2+2�ℎ$

Where:

• $�$ is the final velocity (speed) of the rock just before it hits the ground.
• $�$ is the initial velocity, which is 0 m/s because the rock is dropped from rest.
• $�$ is the acceleration due to gravity, given as $10{\text{m/s}}^2$.
• $ℎ$ is the height from which the rock is dropped, given as $5$ m.

Substituting the known values:

${�}^2=0^2+2\times 10\times 5$ ${�}^2=0+100$ $�=\sqrt{100}$ $�=10\text{m/s}$

So, the speed of the rock just before it hits the ground is $10\text{m/s}$.

Now, to find the kinetic energy of the rock when it reaches the ground, we'll use the formula for kinetic energy ($��$):

$��=\frac{1}{2}�{�}^2$

Where:

• $�$ is the mass of the rock, given as $10$ kg.
• $�$ is the speed of the rock just before it hits the ground, which we found to be $10\text{m/s}$.

Substituting the known values:

$��=\frac{1}{2}\times 10\times (10{)}^2$ $��=\frac{1}{2}\times 10\times 100$ $��=500\text{J}$

So, the kinetic energy of the rock when it reaches the ground is $500\text{J}$.

36. Calculate the work done by the brakes of a car of mass 1000 kg when its speed is reduced from 20 m/s to

10 m/s ?

To calculate the work done by the brakes of the car, we can use the equation:

$�=\frac{1}{2}�({�}_{�}^2-{�}_{�}^2)$

Where:

• $�$ is the work done by the brakes.
• $�$ is the mass of the car, given as $1000$ kg.
• ${�}_{�}$ is the final velocity of the car, given as $10$ m/s.
• ${�}_{�}$ is the initial velocity of the car, given as $20$ m/s.

Substituting the given values into the equation:

$�=\frac{1}{2}\times 1000\times ((10{)}^2-(20{)}^2)$

$�=\frac{1}{2}\times 1000\times (100-400)$

$�=\frac{1}{2}\times 1000\times (-300)$

$�=-150000\text{J}$

The negative sign indicates that work is done by an external force (the brakes) to decrease the kinetic energy of the car. In other words, the brakes exert a force in the direction opposite to the motion of the car, which is why the work done is negative.

So, the work done by the brakes of the car when its speed is reduced from $20\text{m/s}$ to $10\text{m/s}$ is $-150000\text{J}$.

37. A body of mass 100 kg is lifted up by 10 m. Find :

(i) the amount of work done.

(ii) potential energy of the body at that height (value of g = 10 m/s2).

To find the amount of work done and the potential energy of the body at a height of 10 m, we'll use the following formulas:

(i) The amount of work done ($�$) when lifting the body against gravity is given by:

$�=\text{Force}\times \text{Distance}$

Where:

• Force ($�$) = Weight of the body = $��$
• Distance ($�$) = Height lifted

(ii) The potential energy ($��$) of the body at height $ℎ$ is given by:

$��=��ℎ$

Given:

• Mass ($�$) = 100 kg
• Height lifted ($ℎ$) = 10 m
• Acceleration due to gravity ($�$) = 10 m/s²

Let's calculate:

(i) The amount of work done: $�=���$ $�=(100\text{kg})(10{\text{m/s}}^2)(10\text{m})$ $�=10000\text{J}$

So, the amount of work done to lift the body up by 10 m is $10000\text{J}$.

(ii) The potential energy of the body at that height: $��=��ℎ$ $��=(100\text{kg})(10{\text{m/s}}^2)(10\text{m})$ $��=10000\text{J}$

So, the potential energy of the body at a height of 10 m is $10000\text{J}$.

38. A boy weighing 50 kg climbs up a vertical height of 100 m. Calculate the amount of work done by him. How

much potential energy does he gain ? (g = 9.8 m/s2).

To find the amount of work done by the boy and the potential energy he gains, we'll use the following formulas:

1. The amount of work done ($�$) when lifting the boy against gravity is given by:

$�=\text{Force}\times \text{Distance}$

2. The potential energy ($��$) gained by the boy at height $ℎ$ is given by:

$��=��ℎ$

Given:

• Mass of the boy ($�$) = 50 kg
• Height climbed ($ℎ$) = 100 m
• Acceleration due to gravity ($�$) = 9.8 m/s²

Let's calculate:

1. The amount of work done by the boy: $�=\text{Force}\times \text{Distance}$ $�=���$ $�=(50\text{kg})(9.8{\text{m/s}}^2)(100\text{m})$ $�=49000\text{J}$

So, the amount of work done by the boy when climbing up a vertical height of 100 m is $49000\text{J}$.

2. The potential energy gained by the boy: $��=��ℎ$ $��=(50\text{kg})(9.8{\text{m/s}}^2)(100\text{m})$ $��=49000\text{J}$

So, the potential energy gained by the boy at a height of 100 m is $49000\text{J}$.

39. When is the work done by a force on a body : (a) positive, (b) negative, and (c) zero ?

When is the work done by a force on a body: (a) Positive: When the force and displacement are in the same direction. (b) Negative: When the force and displacement are in opposite directions. (c) Zero: When the force and displacement are perpendicular to each other.

40. To what height should a box of mass 150 kg be lifted, so that its potential energy may become 7350 joules ?

(g = 9.8 m/s2).

39. To find the height $ℎ$ at which the potential energy of the box becomes 7350 joules, we use the formula for potential energy ($��$):

$��=��ℎ$

Given $��=7350$ J, $�=150$ kg, and $�=9.8$ m/s², we solve for $ℎ$:

$ℎ=\frac{��}{��}=\frac{7350}{150\times 9.8}$

$ℎ\approx 5\text{ meters}$

So, the box should be lifted to a height of 5 meters.

41. A body of mass 2 kg is thrown vertically upwards with an initial velocity of 20 m/s. What will be its

potential energy at the end of 2 s ? (Assume g = 10 m/s2).

41. To find the potential energy of the body at the end of 2 seconds, we use the formula:

$��=��ℎ$

Given $�=2$ kg, $�=10$ m/s², and $ℎ=\frac{1}{2}�{�}^2$:

$ℎ=\frac{1}{2}�{�}^2=\frac{1}{2}\times 10\times (2{)}^2$

$ℎ=20\text{ meters}$

Now, substitute $ℎ$ into the potential energy formula:

$��=��ℎ=2\times 10\times 20$

$��=400\text{ J}$

So, the potential energy at the end of 2 seconds is $400\text{ J}$.

42. How much work is done when a force of 1 N moves a body through a distance of 1 m in its own direction ?

The work done ($�$) is given by the formula $�=��$. Given $�=1$ N and $�=1$ m, $�=1$ J.

43. A car is being driven by a force of 2.5 × 1010 N. Travelling at a constant speed of 5 m/s, it takes 2 minutes to

reach a certain place. Calculate the work done.

To calculate the work done, we use the formula $�=��$. Given $�=2.5\times 10^{10}$ N, $�=5$ m, and $�=2$ min, we first convert the time to seconds: $�=2\times 60=120$ s. Then, $�=(2.5\times 10^{10})\times 5=1.25\times 10^{11}$ J.

44. Explain by an example that a body may possess energy even when it is not in motion.

Example: A stationary rock on the edge of a cliff possesses potential energy due to its position relative to the ground.

45. (a) On what factors does the gravitational potential energy of a body depend ?

(b) Give one example each of a body possessing : (i) kinetic energy, and (ii) potential energy.

(a) Gravitational potential energy of a body depends on its mass, height, and the acceleration due to gravity. (b) (i) A body possessing kinetic energy: A moving car. (ii) A body possessing potential energy: A book placed on a table.

46. Give two examples where a body possesses both, kinetic energy as well as potential energy.

Examples of bodies possessing both kinetic energy and potential energy: A pendulum at the top of its swing and a roller coaster at the top of a hill.

47. How much is the mass of a man if he has to do 2500 joules of work in climbing a tree 5 m tall ?

(g = 10 m s–2)

46. Given $��=2500$ J and $ℎ=5$ m, we use the formula $��=��ℎ$ to solve for $�$:

$�=\frac{��}{�ℎ}=\frac{2500}{10\times 5}=50\text{ kg}$

So, the mass of the man is 50 kg.

48. If the work done by a force in moving an object through a distance of 20 cm is 24.2 J, what is the magnitude

of the force ?

48. Given $�=24.2$ J and $�=20$ cm, we use the formula $�=��$ to solve for $�$:

$�=\frac{�}{�}=\frac{24.2}{20\times 10^{-2}}=121\text{ N}$

So, the magnitude of the force is 121 N.

49. A boy weighing 40 kg makes a high jump of 1.5 m.

(i) What is his kinetic energy at the highest point ?

(ii) What is his potential energy at the highest point ? (g = 10 m/s2).

(i) Given $�=40$ kg and $�=10$ m/s², the kinetic energy at the highest point is $��=\frac{1}{2}�{�}^2=0\text{J}$ since the velocity is zero. (ii) Given $��=��ℎ=40\times 10\times 1.5=600$ J.

50. What type of energy is possessed :

(a) by the stretched rubber strings of a catapult ?

(b) by the piece of stone which is thrown away on releasing the stretched rubber strings of catapult ?

(a) Stretched rubber strings of a catapult possess elastic potential energy. (b) The piece of stone thrown away on releasing the stretched rubber strings of a catapult possesses kinetic energy.

51. A weightlifter is lifting weights of mass 200 kg up to a height of 2 metres. If g = 9.8 m s–2, calculate :

(a) potential energy acquired by the weights.

(b) work done by the weightlifter.

(a) Given $�=200$ kg and $ℎ=2$ m, we use the formula $��=��ℎ$ to solve for potential energy: $��=200\times 9.8\times 2=3920\text{ J}$ (b) Given $�=��=3920$ J. Since the weightlifter lifts the weights vertically, the work done by the weightlifter is equal to the potential energy acquired by the weights.

Long Answer Type Questions

52. (a) Define the term ‘work’. Write the formula for the work done on a body when a force acts on the body in

the direction of its displacement. Give the meaning of each symbol which occurs in the formula.

(b) A person of mass 50 kg climbs a tower of height 72 metres. Calculate the work done.

(g = 9.8 m s–2 )

(a)

• Definition of Work: In physics, work is defined as the transfer of energy that occurs when a force is applied to an object and causes it to move in the direction of the force. Mathematically, work is equal to the product of the force applied to the object and the displacement of the object in the direction of the force.

• Formula for Work: The formula for work ($�$) done on a body when a force acts on the body in the direction of its displacement is: $�=�\times �$ Where:

  • $�$ is the work done (in joules, J).
  • $�$ is the magnitude of the force applied on the body (in newtons, N).
  • $�$ is the displacement of the body in the direction of the force (in meters, m).

(b) Given:

• Mass of the person ($�$) = 50 kg
• Height of the tower ($ℎ$) = 72 m
• Acceleration due to gravity ($�$) = 9.8 m/s²

The work done ($�$) in lifting the person to the top of the tower against gravity is equal to the change in potential energy of the person:

$�=��ℎ$

Substituting the given values:

$�=(50\text{kg})\times (9.8{\text{m/s}}^2)\times (72\text{m})$

$�=35280\text{J}$

So, the work done by the person in climbing the tower is $35280\text{J}$.

53. (a) When do we say that work is done ? Write the formula for the work done by a body in moving up

against gravity. Give the meaning of each symbol which occurs in it.

(b) How much work is done when a force of 2 N moves a body through a distance of 10 cm in the direction

of force ?

(a)

• When Work is Done: Work is said to be done when a force acts on an object and causes a displacement in the direction of the force. Mathematically, work is done when there is a component of the force in the direction of the displacement.

• Formula for Work Done in Moving Up Against Gravity: The formula for the work done ($�$) by a body in moving up against gravity is given by: $�=��ℎ$ Where:

  • $�$ is the work done (in joules, J).
  • $�$ is the mass of the body (in kilograms, kg).
  • $�$ is the acceleration due to gravity (in meters per second squared, m/s²).
  • $ℎ$ is the height the body is lifted (in meters, m).

This formula calculates the work done when lifting an object vertically against the force of gravity. It is derived from the fact that the work done is equal to the change in potential energy, where potential energy is given by $��=��ℎ$.

(b) Given:

• Force ($�$) = 2 N
• Distance ($�$) = 10 cm = 0.1 m

We'll use the formula $�=��$ to calculate the work done: $�=��=(2\text{N})\times (0.1\text{m})=0.2\text{J}$

So, the work done when a force of 2 N moves a body through a distance of 10 cm in the direction of force is $0.2\text{J}$.

54. (a) What happens to the work done when the displacement of a body is at right angles to the direction of

force acting on it ? Explain your answer.

(b) A force of 50 N acts on a body and moves it a distance of 4 m on a horizontal surface. Calculate the work

done if the direction of force is at an angle of 60° to the horizontal surface.

(a) When the displacement of a body is at right angles to the direction of the force acting on it, the work done by the force is zero. This is because work is defined as the product of force and displacement in the direction of the force. If the displacement is perpendicular to the direction of the force, there is no component of the force acting in the direction of the displacement, resulting in zero work done.

Mathematically, if $�$ represents the angle between the force vector ($�$) and the displacement vector ($�$), the work done ($�$) is given by: $�=��\cos �$

When $�=90^{\circ }$, $\cos �=\cos 90^{\circ }=0$. Therefore, regardless of the magnitude of the force or displacement, if $�=90^{\circ }$, the work done ($�$) will be zero.

(b) Given:

• Force ($�$) = 50 N
• Distance ($�$) = 4 m
• Angle ($�$) = 60°

To calculate the work done, we'll use the formula: $�=��\cos �$

Substituting the given values: $�=(50\text{N})\times (4\text{m})\times \cos 60^{\circ }$

Using $\cos 60^{\circ }=0.5$, we get: $�=(50\text{N})\times (4\text{m})\times 0.5$ $�=100\text{J}$

So, the work done when a force of 50 N acts on a body and moves it a distance of 4 m on a horizontal surface at an angle of 60° to the horizontal surface is $100\text{J}$.

55. (a) Define the term ‘energy’ of a body. What is the SI unit of energy.

(b) What are the various forms of energy ?

(c) Two bodies having equal masses are moving with uniform speeds of v and 2v respectively. Find the ratio

of their kinetic energies.

(a)

• Definition of Energy: Energy is the ability or capacity of a system to do work. It is a scalar quantity that can exist in various forms and can be transferred between objects or converted from one form to another. In simpler terms, energy is the measure of an object's ability to perform work or cause change.
• SI Unit of Energy: The SI unit of energy is the joule (J).

(b)

• Various Forms of Energy: Energy can exist in many forms, including:
  1. Kinetic Energy: Energy associated with the motion of an object.
  2. Potential Energy: Energy associated with the position or configuration of an object within a force field (e.g., gravitational potential energy, elastic potential energy).
  3. Thermal Energy: Energy associated with the temperature of an object or substance.
  4. Chemical Energy: Energy stored in the chemical bonds of molecules.
  5. Electrical Energy: Energy associated with the movement of electrons in a conductor.
  6. Nuclear Energy: Energy stored in the nucleus of an atom and released during nuclear reactions.
  7. Radiant Energy: Energy carried by electromagnetic waves, such as light or radio waves.
  8. Sound Energy: Energy associated with the vibration of particles in a medium, producing sound waves.
  9. Elastic Energy: Energy stored in objects undergoing deformation, such as stretched or compressed springs.
  10. Gravitational Energy: Energy associated with the gravitational attraction between objects in a gravitational field.

(c) Given:

• Masses of both bodies are equal.
• Speed of the first body = $�$
• Speed of the second body = $2�$

To find the ratio of their kinetic energies ($�{�}_1$ and $�{�}_2$), we use the formula for kinetic energy ($��=\frac{1}{2}�{�}^2$), where $�$ is mass and $�$ is velocity:

$�{�}_1=\frac{1}{2}�{�}^2$ $�{�}_2=\frac{1}{2}�(2�{)}^2=2\times \frac{1}{2}�{�}^2=2\times �{�}_1$

So, the ratio of their kinetic energies is: $\frac{�{�}_1}{�{�}_2}=\frac{�{�}_1}{2\times �{�}_1}=\frac{1}{2}$

56. (a) What do you understand by the kinetic energy of a body ?

(b) A body is thrown vertically upwards. Its velocity goes on decreasing. What happens to its kinetic energy

as its velocity becomes zero ?

(c) A horse and a dog are running with the same speed. If the weight of the horse is ten times that of the

dog, what is the ratio of their kinetic energies ?

(a) Kinetic Energy: Kinetic energy is the energy possessed by a body due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Mathematically, kinetic energy ($��$) is calculated using the formula:

$��=\frac{1}{2}�{�}^2$

Where:

• $�$ is the mass of the body.
• $�$ is the velocity of the body.

(b) When a body is thrown vertically upwards and its velocity goes on decreasing, its kinetic energy also decreases. As the velocity of the body becomes zero at the highest point of its trajectory (when it momentarily stops before falling back down), its kinetic energy also becomes zero. At this point, all of the initial kinetic energy is converted into potential energy due to the gravitational force acting on the body.

(c) Given:

• Speed of the horse = Speed of the dog
• Weight of the horse = 10 times weight of the dog

Since kinetic energy depends on the square of velocity and mass, we can find the ratio of their kinetic energies using the formula:

$��=\frac{1}{2}�{�}^2$

As the speed of the horse and the dog is the same, their kinetic energies will be directly proportional to their masses. Since the mass of the horse is ten times that of the dog, the ratio of their kinetic energies will be the same as the ratio of their masses:

$\text{Ratio of kinetic energies}=\frac{\text{Mass of horse}}{\text{Mass of dog}}=\frac{10}{1}=10:1$

So, the ratio of their kinetic energies is 10:1.

57. (a) Explain by an example what is meant by potential energy. Write down the expression for gravitational

potential energy of a body of mass m placed at a height h above the surface of the earth.

(b) What is the difference between potential energy and kinetic energy ?

(c) A ball of mass 0.5 kg slows down from a speed of 5 m/s to that of 3 m/s. Calculate the change in kinetic

energy of the ball. State your answer giving proper units.

(a) Explanation of Potential Energy: Potential energy is the energy stored within an object due to its position or configuration in a force field. One common example of potential energy is gravitational potential energy. When an object is raised to a certain height above the ground, it gains gravitational potential energy because of its position relative to the Earth's surface. This potential energy can be released and converted into kinetic energy when the object falls.

Expression for Gravitational Potential Energy: The expression for gravitational potential energy ($��$) of a body of mass $�$ placed at a height $ℎ$ above the surface of the Earth is given by the formula: $��=��ℎ$ Where:

• $�$ is the mass of the body (in kilograms, kg).
• $�$ is the acceleration due to gravity (in meters per second squared, m/s²).
• $ℎ$ is the height above the surface of the Earth (in meters, m).

(b) Difference between Potential Energy and Kinetic Energy: The main difference between potential energy and kinetic energy lies in their respective forms and causes:

• Potential energy is the energy associated with the position or configuration of an object within a force field. It depends on the object's position or state and its interaction with its surroundings.
• Kinetic energy, on the other hand, is the energy associated with the motion of an object. It depends on the object's mass and velocity.

In simpler terms, potential energy is stored energy due to position, while kinetic energy is the energy of motion.

(c) To calculate the change in kinetic energy ($\mathrm{\Delta }��$) of the ball, we use the formula: $\mathrm{\Delta }��=\frac{1}{2}�({�}_{�}^2-{�}_{�}^2)$ Where:

• $�$ is the mass of the ball (0.5 kg)
• ${�}_{�}$ is the final velocity (3 m/s)
• ${�}_{�}$ is the initial velocity (5 m/s)

Substituting the given values: $\mathrm{\Delta }��=\frac{1}{2}\times 0.5\times ((3{)}^2-(5{)}^2)$ $\mathrm{\Delta }��=\frac{1}{2}\times 0.5\times (9-25)$ $\mathrm{\Delta }��=\frac{1}{2}\times 0.5\times (-16)$ $\mathrm{\Delta }��=-4\text{J}$

So, the change in kinetic energy of the ball is $-4\text{J}$. The negative sign indicates that the kinetic energy decreases as the ball slows down.

58. (a) What is the difference between gravitational potential energy and elastic potential energy ? Give one

example of a body having gravitational potential energy and another having elastic potential energy.

(b) If 784 J of work was done in lifting a 20 kg mass, calculate the height through which it was lifted.

(g = 9.8 m/s2)

(a) Difference between Gravitational Potential Energy and Elastic Potential Energy:

• Gravitational Potential Energy: Gravitational potential energy is the energy stored in an object due to its position in a gravitational field. It arises from the gravitational attraction between the object and the Earth. The gravitational potential energy of an object increases as it is raised to a higher altitude above the Earth's surface and decreases as it moves closer to the Earth's surface. An example of a body having gravitational potential energy is a book placed on a shelf. When the book is lifted higher above the ground, its gravitational potential energy increases.

• Elastic Potential Energy: Elastic potential energy is the energy stored in an object due to its deformation under the action of a force. It arises from the stretching or compressing of an elastic material, such as a spring or a rubber band. The elastic potential energy of an object increases as the material is stretched or compressed further and decreases as the material returns to its original shape. An example of a body having elastic potential energy is a compressed spring. When a spring is compressed, it stores elastic potential energy, which can be released to do work when the spring returns to its original shape.

(b) To calculate the height ($ℎ$) through which the 20 kg mass was lifted, we can use the formula for gravitational potential energy ($��$):

$��=��ℎ$

Given:

• Work done ($�$) = 784 J
• Mass ($�$) = 20 kg
• Acceleration due to gravity ($�$) = 9.8 m/s²

We know that the work done is equal to the change in gravitational potential energy, so we have:

$�=\mathrm{\Delta }��$

Substituting the formula for gravitational potential energy:

$784\text{J}=��ℎ-0$

$784\text{J}=(20\text{kg})(9.8{\text{m/s}}^2)(ℎ)$

$784\text{J}=196ℎ$

$ℎ=\frac{784}{196}\text{m}$

$ℎ=4\text{m}$

So, the height through which the 20 kg mass was lifted is $4\text{m}$.

Multiple Choice Questions (MCQs)

59. A car is accelerated on a levelled road and acquires a velocity 4 times of its initial velocity. During this

process, the potential energy of the car :

(a) does not change (b) becomes twice that of initial potential energy

(c) becomes 4 times that of initial potential energy (d) becomes 16 times that of initial potential energy

(a)

60. A car is accelerated on a levelled road and attains a speed of 4 times its initial speed. In this process, the

kinetic energy of the car :

(a) does not change (b) becomes 4 times that of initial kinetic energy

(c) becomes 8 times that of initial kinetic energy (d) becomes 16 times that of initial kinetic energy

(d)

61. In case of negative work, the angle between the force and displacement is :

(a) 0° (b) 45° (c) 90° (d) 180°

(d)

62. An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass 3.5 kg. Both the

spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the

same :

(a) acceleration (b) momentum (c) potential energy (d) kinetic energy

(a)

63. A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. If the value

of g be 10 m/s2, the work done by the girl against the gravitational force will be :

(a) 6000 J (b) 0.6 J (c) 0 J (d) 6 J

(c)

64. The work done on an object does not depend on the :

(a) displacement (b) angle between force and displacement

(c) force applied (d) initial velocity of the object

(d)

65. Water stored in a dam possesses :

(a) no energy (b) electrical energy (c) kinetic energy (d) potential energy

(d)

66. The momentum of a bullet of mass 20 g fired from a gun is 10 kg.m/s. The kinetic energy of this bullet in

kJ will be :

(a) 5 (b) 1.5 (c) 2.5 (d) 25

(c)

67. Each of the following statement describes a force acting. Which force is causing work to be done ?

(a) the weight of a book at rest on a table

(b) the pull of a moving railway engine on its coaches

(c) the tension in an elastic band wrapped around a parcel

(d) the push of a person’s feet when standing on the floor

(b)

68. A girl weighing 400 N climbs a vertical ladder. If the value of g be 10 m s–2, the work done by her after

climbing 2 m will be :

(a) 200 J (b) 800 J (c) 8000 J (d) 2000 J

(b)

69. Which of the following does not possess the ability to do work not because of motion ? (a) a sparrow flying in the sky (b) a sparrow moving slowly on the ground (c) a sparrow in the nest on a tree (d) a squirrel going up a tree

(c)

70. A stone is thrown upwards as shown in the diagram. When it reaches P, which of the following has the greatest value for the stone ? (a) its acceleration (b) its kinetic energy (c) its potential energy (d) its weight

(c)

Questions Based on High Order Thinking Skills (HOTS)

71. A boy tries to push a truck parked on the roadside. The truck does not move at all. Another boy pushes a

bicycle. The bicycle moves through a certain distance. In which case was the work done more : on the truck

or on the bicycle ? Give a reason to support your answer.

The work done is more in the case of the bicycle being pushed. 

Work is defined as the product of force and displacement in the direction of the force. In both cases, force is applied to move the objects (the truck and the bicycle) through a certain distance. However, the crucial difference lies in the displacement. 

When the first boy pushes the truck and it doesn't move at all, the displacement is zero (since the truck doesn't move). Therefore, despite applying force, no work is done on the truck because the displacement is zero.

In contrast, when the second boy pushes the bicycle and it moves through a certain distance, there is displacement. As a result, work is done on the bicycle because there is both force applied and displacement in the direction of the force.

So, even though both boys exerted force, since the bicycle moved (and thus experienced displacement), work was done on the bicycle, making it the case where more work was done compared to the stationary truck.

72. The work done by a force acting obliquely is given by the formula : W = F cos T × s. What will happen to the

work done if angle T between the direction of force and motion of the body is increased gradually ? Will it

increase, decrease or remain constant ?

The formula given $�=�\cos (�)\times �$ represents the work done by a force acting obliquely, where:

• $�$ is the work done,
• $�$ is the magnitude of the force,
• $�$ (angle $�$ in the question) is the angle between the direction of the force and the direction of motion, and
• $�$ is the displacement.

In this formula, $\cos (�)$ represents the component of the force acting in the direction of motion.

As the angle $�$ between the direction of the force and the motion of the body is increased gradually:

1. When $�=0^{\circ }$ (force acting parallel to motion), $\cos (�)=1$, and maximum work is done.
2. As $�$ increases beyond $0^{\circ }$, the value of $\cos (�)$ gradually decreases. This means the component of the force in the direction of motion decreases.
3. Consequently, as $�$ increases, the work done $�$ decreases because the force applied in the direction of motion (the effective force) decreases.

Therefore, as the angle $�$ increases gradually, the work done will decrease.

73. What should be the angle between the direction of force and the direction of motion of a body so that the

work done is zero ?

The work done is zero when the angle between the direction of force and the direction of motion is $90^{\circ }$. This is because the formula for work done $�=�\cos (�)\times �$ involves the cosine of the angle between the force and the displacement vector.

When the angle $�$ is $90^{\circ }$, $\cos (90^{\circ })=0$, and therefore $�\cos (�)$ becomes zero. This means that no component of the force is acting in the direction of motion, resulting in zero work done on the body.

74. In which of the following case the work done by a force will be maximum : when the angle between the

direction of force and direction of motion is 0° or 90° ?

The work done by a force will be maximum when the angle between the direction of the force and the direction of motion is 0°.

The work done by a force ($�$) is given by the formula:

$�=�\cdot �\cdot \cos (�)$

Where:

• $�$ is the magnitude of the force,
• $�$ is the displacement, and
• $�$ is the angle between the force and the displacement vectors.

When the angle $�$ between the direction of force and the direction of motion is 0°, $\cos (0\mathrm{°})=1$. Therefore, the term $\cos (�)$ becomes its maximum value of 1.

As a result, when the angle $�$ is 0°, the work done ($�$) is maximum. This occurs when the force and the direction of motion are aligned, meaning the force is applied in the same direction as the displacement, leading to maximum work done by the force.

75. How much work is done by the gravitational force of earth acting on a satellite moving around it in a

circular path ? Give reason for your answer.

The gravitational force of Earth does no work on a satellite moving around it in a circular path.

This is because work is defined as the product of force and displacement in the direction of the force. In the case of a satellite moving in a circular orbit around the Earth, the gravitational force always acts perpendicular to the direction of motion (towards the center of the circular path). As a result, there is no displacement in the direction of the force.

According to the definition of work, if the displacement is perpendicular to the force, then the work done is zero. Therefore, the gravitational force of Earth does zero work on the satellite as it orbits the Earth in a circular path.

76. A man is instructed to carry a package from the base camp at B to summit A of a hill at a height of 1200

metres. The man weighs 800 N and the package weighs 200 N. If g = 10 m/s2,

(i) how much work does man do against gravity ?

(ii) what is the potential energy of the package at A if it is assumed to be zero at B ?

(i) The work done by the man against gravity can be calculated using the formula:

$�=�\times �$

Where:

• $�$ is the work done,
• $�$ is the force exerted against gravity, and
• $�$ is the vertical displacement.

The total force exerted against gravity is the sum of the weight of the man and the weight of the package. Since they are both moving vertically upwards, we need to consider their combined weight.

$�=(800\text{N}+200\text{N})=1000\text{N}$

The vertical displacement $�$ is the height of the hill, which is $1200\text{m}$.

$�=1000\text{N}\times 1200\text{m}=1,200,000\text{J}$

Therefore, the man does $1,200,000\text{J}$ of work against gravity.

(ii) The potential energy $�$ of the package at the summit A can be calculated using the formula:

$�=�\times �\times ℎ$

Where:

• $�$ is the mass of the package,
• $�$ is the acceleration due to gravity, and
• $ℎ$ is the height.

Since the package is assumed to be zero potential energy at the base camp, the height $ℎ$ at the summit A is $1200\text{m}$.

$�=200\text{N}\times 10{\text{m/s}}^2\times 1200\text{m}=2,400,000\text{J}$

Therefore, the potential energy of the package at the summit A is $2,400,000\text{J}$.

77. When a ball is thrown vertically upwards, its velocity goes on decreasing. What happens to its potential

energy as its velocity becomes zero ?

As the ball is thrown vertically upwards, its velocity decreases due to the gravitational force acting against its motion. Eventually, the ball reaches its maximum height, where its velocity becomes zero momentarily before it starts to fall back down.

At the point where the ball reaches its maximum height and its velocity becomes zero, its potential energy reaches its maximum value. This is because potential energy is associated with the position of an object relative to a reference point, often defined as the position where the potential energy is zero.

At the highest point of its trajectory, the ball has gained potential energy due to its elevation against the gravitational field of the Earth. This potential energy is converted from the kinetic energy it had while it was moving upwards.

Therefore, as the ball's velocity becomes zero at the highest point of its trajectory, its potential energy is at its maximum.

78. A man X goes to the top of a building by a vertical spiral staircase. Another man Y of the same mass goes to

the top of the same building by a slanting ladder. Which of the two does more work against gravity and

why ?

The amount of work done against gravity depends on the vertical displacement of each person. Let's analyze the situations for both man X and man Y:

1. Man X going up the vertical spiral staircase:

   When man X climbs the vertical spiral staircase, he moves directly against gravity. The staircase is vertical, so the displacement is purely vertical.

   

2. Man Y going up the slanting ladder:

   When man Y climbs the slanting ladder, he not only moves vertically but also horizontally. The displacement here is a combination of vertical and horizontal components. While some work is done against gravity as he ascends vertically, the horizontal component of displacement doesn't contribute to the work done against gravity.

Since both men are of the same mass and reach the same height, the vertical displacement is the same for both. However, man X, climbing the vertical spiral staircase, does work solely against gravity, while man Y, climbing the slanting ladder, does work against both gravity and the horizontal component of displacement.

Therefore, man X does more work against gravity because all his effort is directed vertically, while man Y expends some effort moving horizontally in addition to vertically, which doesn't contribute to the work done against gravity.

79. When a ball is thrown inside a moving bus, does its kinetic energy depend on the speed of the bus ?

Explain.

Yes, the kinetic energy of the ball thrown inside a moving bus does depend on the speed of the bus.

Kinetic energy ($��$) is defined as $\frac{1}{2}�{�}^2$, where $�$ is the mass of the object and $�$ is its velocity.

When the ball is thrown inside the moving bus, its velocity is composed of two parts: its own velocity relative to the ground and the velocity of the bus. The total velocity of the ball is the vector sum of these two velocities.

If we consider the frame of reference of the bus, the ball's velocity relative to the bus is what affects its kinetic energy within that frame. However, when we consider the kinetic energy relative to the ground (which is usually the standard reference frame), we must account for the velocity of the bus as well.

The kinetic energy of an object is directly proportional to the square of its velocity. Therefore, if the bus is moving, its velocity adds to the velocity of the ball relative to the ground. As a result, the kinetic energy of the ball thrown inside a moving bus does depend on the speed of the bus, as the kinetic energy of the ball is affected by the combined velocity of the ball relative to the ground and the velocity of the bus.

80. A bullet of mass 15 g has a speed of 400 m/s. What is its kinetic energy ? If the bullet strikes a thick target

and is brought to rest in 2 cm, calculate the average net force acting on the bullet. What happens to the

kinetic energy originally in the bullet ?

First, let's calculate the kinetic energy ($��$) of the bullet using the formula:

$��=\frac{1}{2}�{�}^2$

Where:

• $�=15\text{g}=0.015\text{kg}$ (mass of the bullet)
• $�=400\text{m/s}$ (speed of the bullet)

$��=\frac{1}{2}\times 0.015\text{kg}\times (400\text{m/s}{)}^2$

$��=\frac{1}{2}\times 0.015\times 160000\text{J}$

$��=1200\text{J}$

Now, let's calculate the average net force ($�$) acting on the bullet using the formula for force:

$�=\frac{\mathrm{\Delta }�}{\mathrm{\Delta }�}$

Where:

• $\mathrm{\Delta }�$ is the change in momentum
• $\mathrm{\Delta }�$ is the time interval over which the change in momentum occurs

The change in momentum ($\mathrm{\Delta }�$) can be calculated using the impulse-momentum theorem:

$\mathrm{\Delta }�=�\mathrm{\Delta }�$

Where:

• $�=0.015\text{kg}$ (mass of the bullet)
• $\mathrm{\Delta }�$ is the change in velocity

Given that the bullet is brought to rest (final velocity is $0$), $\mathrm{\Delta }�=-400\text{m/s}$.

$\mathrm{\Delta }�=0.015\text{kg}\times (-400\text{m/s})$

$\mathrm{\Delta }�=-6\text{Ns}$

Now, we know that the displacement $\mathrm{\Delta }�=2\text{cm}=0.02\text{m}$, and we'll assume the bullet comes to rest uniformly.

$\mathrm{\Delta }�=\frac{\mathrm{\Delta }�}{{�}_{�}}$

Where ${�}_{�}=0$ (final velocity).

$\mathrm{\Delta }�=\frac{0.02\text{m}}{0\text{m/s}}$

$\mathrm{\Delta }�=\text{undefined}$

Since the final velocity is $0$, we cannot use the formula $�=\frac{\mathrm{\Delta }�}{\mathrm{\Delta }�}$ directly because it results in division by zero. Instead, we can calculate the average force using the formula:

${�}_{\text{avg}}=\frac{\mathrm{\Delta }�}{\mathrm{\Delta }�}$

${�}_{\text{avg}}=\frac{-6\text{Ns}}{\text{undefined}}$

${�}_{\text{avg}}=\text{undefined}$

However, if we consider the process of bringing the bullet to rest to be an impulsive force acting over a very short time, we can approximate the average force as very large.

Now, regarding what happens to the kinetic energy originally in the bullet: The kinetic energy of the bullet is converted into other forms of energy during the collision with the target. This includes sound, heat, and deformation energy of the target material. Since the bullet comes to rest, all of its initial kinetic energy is dissipated during the collision.