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CBSE · Class 11 · Physics

Work, Energy and Power

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
A force of 18 N moves a body 11 m, at 180° to the direction of the force. The work done is:
  1. 99 J
  2. 198 J
  3. −199 J
  4. −198 J
2.
The kinetic energy of a 19 kg body moving at 19 m/s is:
  1. 3,429.5 J
  2. 6859 J
  3. 180.5 J
  4. 1,714.75 J
3.
A machine does 4000 J of work in 7 s. Its power is:
  1. 0.00175 W
  2. 9.52 W
  3. 28000 W
  4. 571.43 W
4.
A spring of spring constant 1450 N/m is stretched by 1 cm. The energy stored is:
  1. 0.0725 J
  2. 14.5 J
  3. 0.145 J
  4. 7.25 J
5.
The work done by a force perpendicular to the displacement is:
  1. zero
  2. negative
  3. equal to Fs
  4. maximum
6.
1,800 J of work is done on a 4 kg body at rest. Its speed becomes:
  1. 32 m/s
  2. 30 m/s
  3. 60 m/s
  4. 450 m/s
7.
A 6 kg body moving at 19 m/s collides with a 2 kg body at rest, and they stick together. Their common speed is:
  1. 14.25 m/s
  2. 19 m/s
  3. 9.5 m/s
  4. 4.75 m/s
8.
Assertion (A): If the speed of a 5 kg body increases from 12 m/s to 36 m/s, its kinetic energy increases from 360 J to 3,240 J.
Reason (R): Kinetic energy is proportional to the square of the speed.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
A 12 kg ball falls from rest through a height of 45 m. Find its potential energy at the top and its speed just before it hits the ground (ignore air resistance). (Take g = 10 m/s2.)
10.
A 16 kg box is pulled 15 m across a floor by a horizontal force of 33 N. The coefficient of friction is 0.1. Find the work done by the applied force, by friction, and the gain in kinetic energy. (Take g = 10 m/s2.)
11.
A pump lifts 2900 kg of water through a height of 40 m in 10 minutes. Find the power of the pump (in W and kW). (Take g = 10 m/s2.)
12.
A 2 kg ball moving at 18 m/s makes a head-on elastic collision with a 4 kg ball at rest. Find their velocities after the collision.
13.
A 10 kg trolley moving at 14 m/s hits a 8 kg trolley at rest, and they move off together. Find the common speed and the loss of kinetic energy.
14.
A 4 kg block moving at 10 m/s on a smooth floor hits a spring of spring constant 64 N/m. Find the greatest compression of the spring.
15.
A motor takes in 3500 J of electrical energy and lifts a load, doing 2,625 J of useful work. Find its efficiency and the energy wasted.
16.
A force increases steadily from 0 to 90 N over the first 1 m, then stays at 90 N for the next 5 m. Find the work done (area under the F–x graph).
Section C — Word problems4 × 4 = 16 marks
17.
A 4 kg cart moves at 0 m/s at the top of a frictionless track 170 m high and runs down to a point 45 m high. Find (a) its loss of potential energy, (b) its kinetic energy at the lower point, (c) its speed there (2 decimal places if needed). (Take g = 10 m/s2.)
18.
A bullet of mass 50 g moving at 443 m/s embeds itself in a 5 kg block hanging from a string. Find (a) the speed of the block just after impact, (b) the height it rises through. (2 decimal places.) (Take g = 10 m/s2.)
19.
A car of mass 1700 kg has an engine of power 23 kW. Its greatest speed on a level road is 42 m/s. (a) What is the resistance to motion at that speed? (b) If the resistance stays the same, what is the car's acceleration when it moves at 32 m/s at full power?
20.
Two balls of 4 kg and 2 kg move towards each other at 9 m/s and 4 m/s and collide elastically head-on. Find their velocities after the collision (take the first ball's direction as positive), and check that kinetic energy is conserved.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 11 · Physics

Answer key

Work, Energy and Power · AW-11P-WEP-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) −198 JW = Fs cos θ = 18 × 11 × −1
2(a) 3,429.5 JKE = ½mv2
3(d) 571.43 WP = W/t
4(a) 0.0725 JU = ½kx2
5(a) zero—
6(b) 30 m/sW = ½mv2
7(a) 14.25 m/sm1u1 = (m1 + m2)v
8(a) Both A and R are true, and R is the correct explanation of A.A is true: ½ × 5 × 362 = 3,240 J. R is true and explains A.
95,400 J; 30 m/smgh = ½mv2
10495 J; −240 J; 255 Jwork–energy theorem: net work = change in KE
111,933.33 W = 1.93 kWP = mgh/t = 2900 × 10 × 40 ÷ 600
12−6 m/s and 12 m/sv1 = (m1 − m2)u1/(m1 + m2); v2 = 2m1u1/(m1 + m2)
137.78 m/s; 435.56 JKE before 980 J; after 544.44 J
142.5 m½mv2 = ½kx2, so x = v√(m/k)
1575%; 875 Jefficiency = useful output ÷ input × 100
16495 Jtriangle ½ × 1 × 90 + rectangle 5 × 90
17(a) 5,000 J; (b) 5,000 J; (c) 50 m/senergy is conserved: ½mv12 + mgh1 = ½mv22 + mgh2
18(a) 4.39 m/s; (b) 0.96 mmomentum: 0.05 × 443 = (0.05 + 5)V; then ½V2 = gh
19(a) 547.62 N; (b) 0.10 m/s2at top speed P = Rv; at lower speed driving force = P/v
200.33 m/s and 13.33 m/s; KE before = 178 J = KE afterv1 = [(m1 − m2)u1 + 2m2u2]/(m1 + m2); v2 = [(m2 − m1)u2 + 2m1u1]/(m1 + m2)