This set of Class 11 Physics Chapter 6 Multiple Choice Questions & Answers (MCQs) focuses on “Work-Energy Theorem”.

1. According to the work-energy theorem, total change in energy is equal to the _______

a) Total work done

b) Half of the total work done

c) Total work done added with frictional losses

d) Square of the total work done

View Answer

Explanation: According to the work-energy theorem, the total change in the energy of the system is equal to the total work done. Losses due to friction are also a part of the total work. The total work done is the sum of the work done by all the forces.

2. The work done by a body while covering a vertical height of 5m is 50 kJ. By how much amount has the energy of the body changed?

a) 50 kJ

b) -50 kJ

c) 25 kJ

d) 10 kJ

View Answer

Explanation: The work done by the body is 50 kJ. Hence by the work-energy theorem, we can say that the total change in energy is equal to the total work done. Therefore, the total change in energy is 50 kJ. Since only the ‘amount’ has been asked, the sign, which denotes the direction of energy transfer, does not matter.

3. The total energy change of a body which covers a vertical height is constituted completely by the potential energy change only when ______

a) It moves slowly

b) It moves quickly

c) It moves for some time, then rests, then moves again

d) It moves with a decreasing velocity

View Answer

Explanation: The total change in energy is constituted by the changes in potential and kinetic energies. When the body is moving slowly, it implies that the body is moving with such a small velocity that the square of the velocity in the expression of kinetic energy becomes negligible. Hence, the total change in the energy is constituted by the total change in the potential energy.

4. The change in the total energy of a body moving because of work being done on it is attributed by the change in kinetic energy only when ______

a) It moves extremely slowly

b) It moves with a very high velocity

c) It moves on an equipotential surface

d) It moves with a positive acceleration

View Answer

Explanation: An equipotential surface is defined as a surface on which the potential of the body is same everywhere. Hence the potential of a body moving on an equipotential surface does not change. Therefore, the total change in energy is attributed by the change in the kinetic energy of the body.

5. A body having 7 kg of mass moves with a velocity of 20 m/s covers a vertical height of 8.75 m. What is the work done by the moving force in Joules? Take g = 10 m/s^{2}.

a) 0

b) 10

c) 15

d) 20

View Answer

Explanation: By work-energy theorem, the total work done on a body is equal to the change in the total energy. K.E. = \(\Big(\frac{1}{2}\Big)\)mv

^{2}and P.E. = mgh. One can find out the total change in the kinetic and potential energies at the initial and the final points, i.e. at the starting and after covering a distance of 8.75 m. The total change in the total energy is 0. Hence the work done is 0.

6. A projectile has both change in kinetic and potential energy. Still the work done calculated between any two points is zero. This is due to _________

a) Presence of no external force

b) No displacement

c) The displacement being in both vertical and horizontal directions

d) Inter-conversion of kinetic and potential energies

View Answer

Explanation: When a projectile moves, there is an inter-conversion of kinetic and potential energies. This is what causes the body to cover a maximum vertical and horizontal distance. One might argue over the presence of gravity which is an external force. Well, the work done by gravity is nothing but the potential energy of the body which has already been taken care of in the expression for total energy.

7. A body having 7 kg of mass moves slowly and covers a vertical height of 14 m. What is the work done in Joules and what force causes this work? Take g = 10 m/s^{2}.

a) 980, gravity

b) 980, force acting opposite to gravity

c) 1000, gravity

d) 1000, force acting opposite to gravity

View Answer

Explanation: By work-energy theorem, the total work done on a body is equal to the change in the total energy. Since the body is moving slowly, the change in kinetic energy is negligible. P.E. = mgh. The change in potential energy and hence in the total energy is 980 J. Since only gravity is acting on the body, gravity is what causes the work.

8. The total energy of a system moving under the action of a conservative force changes by 50 J. Which of the following statements are correct with regards to the amount of work done?

a) Amount of work done = 50 J

b) Amount of work done < 50 J

c) Amount of work done > 50 J

d) Amount of work done <= 50 J

View Answer

Explanation: By work-energy theorem, the total work done on a body is equal to the change in the total energy. No matter what kind of force is acting on the system, the theorem upholds. Since the total change in the energy is of 50 J, the total work done is equal to 50 J.

9. A body of mass 2 Kg, initially at rest, moves under the influence of an external force of magnitude 4 N on plane ground. What is the work done by the force and the change in kinetic energy in the first 10 seconds?

a) 400 J, 400 J

b) -400 J, 400 J

c) 400 J, -400 J

d) -400 J, -400 J

View Answer

Explanation: By work-energy theorem, the total work done on a body is equal to the change in the total energy. K.E. = \(\Big(\frac{1}{2}\Big)\)mv

^{2}. The acceleration of the body is 4/2 = 2 m/s

^{2}. The final velocity after 10 s can be found out by the first equation of motion. The total change in the total energy is the total change in kinetic energy as the body is moving on plane ground and is equal to 400 J. Hence the work done is 400 J.

10. A body moving with a constant velocity on normal ground initially has an energy of 1000 J. If the final energy is 800 J, what is the work done and which force causes this work?

a) 200 J, gravity

b) -200 J, friction

c) 200 J, friction

d) -200 J, gravity

View Answer

Explanation: According to the work-energy theorem, the total change in energy is equal to the total work done. The change in energy is -200 J. Hence the work done is -200 J. Since it is mentioned that the body is moving on a normal ground, there is force is friction. Therefore, friction is what causes this work.

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