Questions & Answers: "Work, Energy, and Power"

Complete guide to "Work, Energy, and Power" for Physics students. Below you will find important questions and model answers to help you prepare.

4 Questions

Previous Question Papers

Top

Question 1

1 Mark

What wlll be the power developed by a gun when Soldier fires a bullet of mass 20 g at the rate of 1 bullet/sec with a velocity of 1000 m/s ?

Options

Option A

100 W

Option B

1000 W

Option C is correct

10 kW

Option D

100 kW

General Practice

Top

Question 1

1 Mark

A person lifts a 2 kg block from the ground to a height of 5 meters. Calculate the work done by the person against gravity. (Assume g=10 m/s2g = 10 \ m/s^2)

Options

Option A

0 J

Option B

10 J

Option C is correct

100 J

Option D

200 J

Explanation

The work done by the person against gravity is equal to the potential energy gained by the block. The formula for potential energy is PE=mghPE = mgh, where mm is the mass, gg is the acceleration due to gravity, and hh is the height. Given: Mass (mm) = 2 kg Height (hh) = 5 m Acceleration due to gravity (gg) = 10 m/s210 \ m/s^2 Work done (WW) = PE=mgh=2×10×5=100 JPE = mgh = 2 \times 10 \times 5 = 100 \ J. Therefore, the work done by the person against gravity is 100 J. Options A, B, and D are incorrect as they do not match the calculated work done.

Question 2

1 Mark

A simple pendulum of mass 200 g is released from rest at a height of 0.2 m above its lowest point. Assuming no energy loss due to air resistance, what will be the speed of the pendulum bob when it passes through its lowest point? (Take g=10 m/s2g = 10 \ m/s^2)

Options

Option A

1 m/s

Option B is correct

2 m/s

Option C

4 m/s

Option D

0.5 m/s

Explanation

According to the Law of Conservation of Energy, the potential energy (PE) at the highest point is converted entirely into kinetic energy (KE) at the lowest point, assuming no energy loss. Potential Energy at the highest point (PEtopPE_{top}) = mghmgh Kinetic Energy at the lowest point (KEbottomKE_{bottom}) = 12mv2\frac{1}{2}mv^2 By Conservation of Energy: PEtop=KEbottomPE_{top} = KE_{bottom} mgh=12mv2mgh = \frac{1}{2}mv^2 We can cancel 'm' from both sides: gh=12v2gh = \frac{1}{2}v^2 Rearranging for vv: v2=2gh    v=2ghv^2 = 2gh \implies v = \sqrt{2gh} Given: Height (hh) = 0.2 m Acceleration due to gravity (gg) = 10 m/s210 \ m/s^2 v=2×10×0.2=4=2 m/sv = \sqrt{2 \times 10 \times 0.2} = \sqrt{4} = 2 \ m/s. Therefore, the speed of the pendulum bob at its lowest point is 2 m/s. Options A, C, and D are incorrect based on the conservation of energy principle and calculation.

Question 3

1 Mark

A pump lifts 500 kg of water from a depth of 20 m to the surface in 10 seconds. What is the power of the pump? (Take g=10 m/s2g = 10 \ m/s^2)

Options

Option A

1000 W

Option B

5000 W

Option C is correct

10000 W

Option D

20000 W

Explanation

Power is defined as the rate at which work is done or energy is transferred. First, calculate the work done by the pump to lift the water. The work done against gravity is equal to the potential energy gained by the water. Work done (WW) = mghmgh Given: Mass (mm) = 500 kg Height (hh) = 20 m Acceleration due to gravity (gg) = 10 m/s210 \ m/s^2 W=500×10×20=100000 JW = 500 \times 10 \times 20 = 100000 \ J. Next, calculate the power using the formula: Power (PP) = Work doneTime\frac{Work \ done}{Time} Given: Time (tt) = 10 s P=100000 J10 s=10000 WP = \frac{100000 \ J}{10 \ s} = 10000 \ W. Therefore, the power of the pump is 10000 W. Options A, B, and D are incorrect as they do not match the calculated power.