Free Work and Energy MCQs with Answers

1. Work done by a force is zero when the angle between the force and the displacement is

• A. 0 degrees
• B. 45 degrees
• C. 90 degrees
• D. 180 degrees

Explanation: Work is the scalar product Fd cos theta, and cos 90 degrees is zero, so a force perpendicular to the displacement does no work. This is why the centripetal force in uniform circular motion does no work and the speed stays constant. At 180 degrees the work is negative rather than zero, as with friction.

2. A porter carries a suitcase horizontally across a platform at constant speed. The work done by the porter against gravity is

• A. equal to mgh
• B. zero
• C. equal to the weight times the distance walked
• D. negative

Explanation: The upward force supporting the suitcase is perpendicular to the horizontal displacement, so the work done against gravity is zero. The porter still expends metabolic energy holding the case up, but that is not mechanical work on the suitcase, which is the standard point of this question. Work against gravity appears only when the height changes.

3. The kinetic energy of a body is doubled when its

• A. mass is halved
• B. speed is doubled
• C. speed is increased by a factor of the square root of two
• D. mass is doubled and speed halved

Explanation: Kinetic energy is half m v squared, so it scales with the square of the speed. Multiplying the speed by root 2 multiplies the energy by 2, whereas doubling the speed multiplies the energy by 4. The last option leaves energy at half its original value, since doubling m and quartering v squared gives a net factor of one half.

4. A body of mass 5 kg moves at 4 m s-1. Its kinetic energy is

• A. 10 J
• B. 20 J
• C. 40 J
• D. 80 J

Explanation: Kinetic energy is half times 5 times 4 squared, which is half times 5 times 16, giving 40 J. The common slip is to forget to square the velocity, which gives 10 J, or to forget the factor of one half, which gives 80 J.

5. The work-energy theorem states that the net work done on a body equals

• A. its total mechanical energy
• B. the change in its kinetic energy
• C. the change in its potential energy
• D. its momentum

Explanation: The work-energy theorem equates net work with the change in kinetic energy, which is why a braking force doing negative work brings a car to rest. It holds whether the force is constant or varying. Changes in potential energy relate to the work done by conservative forces specifically, which is a narrower statement.

6. Power is correctly defined as

• A. the total work done
• B. the rate of doing work
• C. work done times time taken
• D. force times displacement

Explanation: Power is work done per unit time, measured in watts, where one watt is one joule per second. For a body moving at constant velocity it can also be written as force times velocity. Force times displacement is work itself, not the rate at which it is done.

7. A pump raises 200 kg of water through 10 m in 20 s. Taking g as 10 m s-2, its output power is

• A. 100 W
• B. 1000 W
• C. 2000 W
• D. 20000 W

Explanation: The work done is mgh, which is 200 times 10 times 10, giving 20000 J. Dividing by the 20 s taken gives 1000 W, or 1 kW. Forgetting to divide by time leaves the answer at 20000, which is the energy in joules rather than the power in watts.

8. A ball is dropped from rest. Ignoring air resistance, as it falls its

• A. total mechanical energy increases
• B. potential energy converts into kinetic energy with the total unchanged
• C. kinetic energy converts into potential energy
• D. total mechanical energy decreases

Explanation: Gravity is a conservative force, so in the absence of air resistance the sum of kinetic and potential energy stays constant while potential energy is steadily converted into kinetic energy. With air resistance included the total mechanical energy would fall, the loss appearing as heat. Energy is never created, so the total cannot increase.

9. A machine takes in 500 J and delivers 350 J of useful output. Its efficiency is

• A. 30%
• B. 43%
• C. 70%
• D. 143%

Explanation: Efficiency is useful output divided by total input times 100, which is 350 over 500 times 100, giving 70%. The remaining 150 J is lost, mostly as heat through friction. An efficiency above 100% is impossible, so the last option can be discarded on principle alone.

10. The gravitational potential energy gained by a 2 kg mass lifted through 5 m, taking g as 10 m s-2, is

• A. 10 J
• B. 25 J
• C. 100 J
• D. 1000 J

Explanation: Potential energy gained is mgh, which is 2 times 10 times 5, giving 100 J. The route taken to lift the mass makes no difference, since gravity is a conservative force and only the change in height counts. The answer 10 J comes from omitting the height.