Work and Energy Class 9 Numericals with Solutions

πŸ“… Published on: 13.12.2025

Work and Energy Class 9 Numericals with Solutions

Work and Energy is one of the most important chapters in Class 9 Physics because it introduces students to fundamental concepts like work, power, kinetic energy, and potential energy through practical numerical problems. Solving Work and Energy Class 9 numericals with solutions helps students clearly understand formulas, units, and real-life applications of energy transfer. These numericals strengthen problem-solving skills, improve accuracy in calculations, and build confidence for school exams as well as competitive tests. Regular practice of solved numericals also helps students avoid common mistakes and develop a strong conceptual base for higher classes.

WhatsApp Group
Join Now
Telegram Group
Join Now
Facebook Profile
Open
YouTube Channel
Subscribe

Work, Energy and Power – All Formulae (Class 9)

1. Work

W = F Γ— s
W = F Γ— s Γ— cosΞΈ
  • If displacement = 0 β†’ Work = 0
  • If force βŸ‚ displacement β†’ Work = 0

2. Energy

  • Energy = Capacity to do work
  • SI unit: Joule (J)

3. Kinetic Energy (KE)

KE = Β½mvΒ²
Work done = Change in KE
W = Β½m(vΒ² βˆ’ uΒ²)
KE = pΒ² / 2m

4. Potential Energy (PE)

PE = mgh
Ξ”PE = mg(hβ‚‚ βˆ’ h₁)

5. Mechanical Energy

Mechanical Energy = KE + PE

6. Law of Conservation of Energy

KE₁ + PE₁ = KEβ‚‚ + PEβ‚‚
  • Total energy remains constant

7. Power

P = W / t
P = F Γ— v
1 Watt = 1 Joule / second

8. Commercial Unit of Energy

1 kWh = 3.6 Γ— 10⁢ J

9. Motion Formulae Used

v = u + at
vΒ² = uΒ² + 2as
s = ut + Β½atΒ²

10. Important Free-Fall Relations

  • At highest point β†’ KE = 0, PE = maximum
  • At ground β†’ PE = 0, KE = maximum
  • During fall β†’ KE + PE = constant

11. SI Units Table

QuantitySI Unit
WorkJoule (J)
EnergyJoule (J)
PowerWatt (W)
ForceNewton (N)
Masskg
Velocitym/s

Work and Energy Class 9 Numericals with Solutions

Class 9 – Work and Energy (Numericals)

Q1. A force of 10 N causes a displacement of 2 m in a body in its own direction. Calculate the work done by force.

Given:
Force (F) = 10 N
Displacement (s) = 2 m

Formula:
W = F Γ— s

Calculation:
W = 10 Γ— 2 = 20 J

Answer: 20 J

Q2. How much force is applied on the body when 150 J of work is done in displacing the body through a distance of 10 m in the direction of force?

Given:
Work (W) = 150 J
Displacement (s) = 10 m

Formula:
W = F Γ— s

Calculation:
F = W Γ· s = 150 Γ· 10 = 15 N

Answer: 15 N

Q3. An engine works 54,000 J by exerting a force of 6000 N. What is the displacement of the force?

Given:
Work (W) = 54,000 J
Force (F) = 6000 N

Formula:
W = F Γ— s

Kinetic Energy – Questions with Answers (Class 9)

Q4. A moving body of 30 kg has 60 J of kinetic energy. Calculate the speed.

KE = Β½mvΒ² 60 = Β½ Γ— 30 Γ— vΒ² vΒ² = 4 v = 2 m/s

Q5. A body of mass 2 kg is moving with a speed of 20 m/s. Find its kinetic energy.

KE = Β½ Γ— 2 Γ— 20Β² KE = 400 J

Q6. A hammer of mass 1 kg falls freely from a height of 2 m. Calculate velocity and KE just before touching the ground.

v = √(2gh) = √(2 Γ— 9.8 Γ— 2) = 6.29 m/s
KE = Β½ Γ— 1 Γ— (6.29)Β² = 19.6 J
Velocity does not depend on mass.

Q7. What change is required in velocity to keep KE same if mass becomes 4 times?

To keep KE constant, velocity becomes half of the original velocity.

Q8. Calculate the work required to stop a car of mass 1500 kg moving at 60 km/h.

v = 16.67 m/s KE = Β½ Γ— 1500 Γ— (16.67)Β² Work = 208333.3 J

Q9. Work done to increase speed of a car from 30 km/h to 60 km/h (mass = 1500 kg).

W = Β½m(vΒ² βˆ’ uΒ²) W = 156250 J

Q10. KE of a body is 25 J at 5 m/s. Find KE when velocity is doubled and tripled.

Velocity doubled β†’ KE = 100 J
Velocity tripled β†’ KE = 225 J

Q11. A force changes velocity of a 20 kg mass from 5 m/s to 2 m/s. Find work done.

W = Β½m(vΒ² βˆ’ uΒ²) W = βˆ’210 J

Q12. Work required to stop a car of mass 1500 kg moving at 60 km/h.

Work = 208333.3 J

Q13. How fast should a 60 kg man run to have KE of 750 J?

Speed = 5 m/s

Q14. Find mass of a body having KE = 5 J at speed 2 m/s.

Mass = 2.5 kg

Q15. A ball of mass 250 g is kicked at 10 m/s. Find work done.

Work done = 12.5 J

Q16. A 5 kg body is acted upon by a force of 20 N for 10 s. Find KE.

Acceleration = 4 m/sΒ² Final velocity = 40 m/s KE = 4000 J

Q17. A 20 g bullet enters a tree at 500 m/s and exits at 400 m/s. Find work done.

Work done = βˆ’900 J

Q18. Find KE of a 15 kg object moving at 4 m/s.

KE = 120 J

Class 9 – Kinetic Energy Numericals (Solved)

Q19. A bullet of mass 0.03 kg moving with a velocity of 400 m/s penetrates 12 cm into a wooden block. Find the resistive force and initial kinetic energy.

KE = Β½mvΒ² = Β½ Γ— 0.03 Γ— 400Β² = 2400 J
Work done = Force Γ— distance
2400 = F Γ— 0.12
F = 20000 N

Q20. Two bodies of equal masses move with velocities v and 3v. Find ratio of KE.

KE ∝ v²
Ratio = vΒ² : (3v)Β² = 1 : 9
Answer: 1 : 9

Q21. Mass of ball A is double of B. A moves with half the speed of B. Find ratio of KE.

KE ∝ mv²
A : B = (2m Γ— (v/2)Β²) : (m Γ— vΒ²) = 1 : 2
Answer: 1 : 2

Q22. A truck weighing 5000 kgf and a cart weighing 500 kgf move with same speed. Compare KE.

KE ∝ mass
Ratio = 5000 : 500 = 10 : 1
Answer: 10 : 1

Q23. A bullet of mass 20 g passes two points 30 m apart in 4 s. Find KE.

Speed = 30 Γ· 4 = 7.5 m/s
Mass = 0.02 kg
KE = Β½ Γ— 0.02 Γ— 7.5Β² = 0.5625 J

Q24. How fast should a man of 50 kg run to have KE = 625 J?

625 = Β½ Γ— 50 Γ— vΒ²
vΒ² = 25 β†’ v = 5 m/s

Q25. Find KE of a body of mass 1 kg moving at 2 m/s.

KE = Β½ Γ— 1 Γ— 2Β² = 2 J

Q26. Find momentum of a body of mass 100 g having KE = 20 J.

KE = pΒ² / 2m
p = √(2mKE) = √(2 Γ— 0.1 Γ— 20)
p = 2 kgΒ·m/s

Q27. Two equal masses move with speeds 2 m/s and 6 m/s. Find ratio of KE.

KE ∝ v²
Ratio = 2Β² : 6Β² = 4 : 36 = 1 : 9

Q28. A 2 kg body falls from rest. Find KE after 2 s (g = 10 m/sΒ²).

v = gt = 10 Γ— 2 = 20 m/s
KE = Β½ Γ— 2 Γ— 20Β² = 400 J

Q29. A scooter slows from 10 m/s to 5 m/s. Mass = 150 kg. Find work done by brakes.

W = Β½m(vΒ² βˆ’ uΒ²)
= Β½ Γ— 150 Γ— (25 βˆ’ 100)
W = βˆ’5625 J

Potential Energy & Law of Conservation of Energy – Class 9

Q30. A body of 5 kg is raised to a height of 2 m. Find the work done.

Work done = mgh
= 5 Γ— 9.8 Γ— 2
Work done = 98 J

Q31. If g = 10 m/sΒ², find the potential energy of a 1 kg body kept at a height of 5 m.

PE = mgh = 1 Γ— 10 Γ— 5
Potential Energy = 50 J

Q32. A work of 4900 J is done to lift a 50 kg load. Find the height.

W = mgh
4900 = 50 Γ— 9.8 Γ— h
h = 10 m
Height = 10 m

Q33. A bag of wheat of mass 200 kg has PE = 9800 J. Find the height (g = 9.8 m/sΒ²).

9800 = 200 Γ— 9.8 Γ— h
h = 5 m
Height = 5 m

Q34. Two bodies of equal masses are kept at heights h and 2h. Find ratio of their potential energies.

PE ∝ height
Ratio = h : 2h = 1 : 2

Q35. Find the energy of a stone of mass 10 kg kept at a height of 5 m.

PE = 10 Γ— 9.8 Γ— 5
Potential Energy = 490 J

Q36. If 196 Γ— 10Β² J energy is used to raise a 40 kg boy, find the height.

Energy = 19600 J
19600 = 40 Γ— 9.8 Γ— h
h = 50 m
Height = 50 m

Q37. Work done to lift a 10 kg body is 490 J. Find the height.

490 = 10 Γ— 9.8 Γ— h
h = 5 m
Height = 5 m

Q38. A 4 kg body is moved from 5 m to 10 m height. Find increase in potential energy.

Ξ”PE = mg(hβ‚‚ βˆ’ h₁)
= 4 Γ— 9.8 Γ— (10 βˆ’ 5)
Increase in PE = 196 J

Q39. A 1 kg object is raised through height h. Its PE increases by 1 J. Find h.

PE = mgh
1 = 1 Γ— 9.8 Γ— h
h = 1/9.8 m

Q40. A 5 kg ball is thrown upwards with speed 10 m/s. Find (a) PE at highest point (b) maximum height.

KE = Β½mvΒ² = Β½ Γ— 5 Γ— 10Β² = 250 J
At highest point KE = 0 β‡’ PE = 250 J
h = PE / mg = 250 / (5 Γ— 10) = 5 m

Q41. A 5 kg ball is dropped from height 10 m. Find PE, KE before ground, velocity.

Initial PE = 5 Γ— 9.8 Γ— 10 = 490 J
KE before ground = 490 J
v = √(2gh) = √(196) = 14 m/s

Q42. A body is thrown up with KE 10 J and reaches height 5 m. Find mass.

KE = PE β‡’ 10 = m Γ— 10 Γ— 5
m = 0.2 kg

Q43. A rocket of mass 3Γ—10⁢ kg reaches height 25 km with speed 1 km/s. Find PE and KE.

PE = mgh = 3Γ—10⁢ Γ— 10 Γ— 25Γ—10Β³
PE = βˆ’7.5 Γ— 10ΒΉΒΉ J
KE = Β½mvΒ² = Β½ Γ— 3Γ—10⁢ Γ— (1000)Β²
KE = 1.5 Γ— 10ΒΉΒ² J

Q44. Find energy of a 10 kg object at height 6 m (g = 9.8).

PE = 10 Γ— 9.8 Γ— 6 = 588 J

Q45. A 5 kg body falls from 5 m. How much energy does it possess?

Total energy = mgh = 5 Γ— 9.8 Γ— 5 = 245 J

Q46. A 12 kg object has PE 480 J. Find height (g = 10).

480 = 12 Γ— 10 Γ— h
h = 4 m

Q47. Find increase in PE when a 2 kg block is lifted through 2 m.

Ξ”PE = 2 Γ— 9.8 Γ— 2 = 39.2 J

Q48. A 1 kg ball is dropped from 5 m. Find KE before ground and speed.

KE = mgh = 49 J
v = √(2gh) = √(98) = 9.87 m/s

work and energy and power class 9 numericals with solutions

Work, Energy & Power – Class 9 (Solved Numericals)

Q51. A spring is compressed by a 150 g toy cart. On release it moves with speed 0.2 m/s. Find elastic PE.

KE = Β½mvΒ² = Β½ Γ— 0.15 Γ— (0.2)Β²
Elastic PE = 0.003 J

Q52. A 40 kg object is raised to 5 m. Find PE and KE at half-way down.

PE = mgh = 40 Γ— 9.8 Γ— 5 = 1960 J
At half height, KE = 980 J

Q53. A 5.8 kg box gains 145 J PE. Find height (g = 10).

145 = 5.8 Γ— 10 Γ— h
h = 2.5 m

Q54. A man gains 2268 J PE climbing a 3.6 m wall. Find mass (g = 10).

2268 = m Γ— 10 Γ— 3.6
m = 63 kg

Q55. A 15 g bullet moves at 400 m/s and stops in 2 cm. Find KE and average force.

KE = Β½ Γ— 0.015 Γ— 400Β² = 1200 J
F = W/s = 1200 / 0.02 = 60000 N
KE converts into heat and sound.

Q56. A 200 g ball falls from 5 m. Find KE on reaching ground.

KE = mgh = 0.2 Γ— 9.8 Γ— 5 = 9.8 J

Q57. A 10 kg rock falls from 5 m. Find speed and KE (g = 10).

v = √(2gh) = √100 = 10 m/s
KE = Β½ Γ— 10 Γ— 10Β² = 500 J

Q58. Find work done by brakes when a 1000 kg car slows from 20 m/s to 10 m/s.

W = Β½m(vΒ² βˆ’ uΒ²)
= Β½ Γ— 1000 Γ— (100 βˆ’ 400)
W = βˆ’150000 J

Q59. A 100 kg body is lifted by 10 m. Find work done and PE (g = 10).

Work = mgh = 100 Γ— 10 Γ— 10 = 10000 J
PE = 10000 J

Q60. A 50 kg boy climbs 100 m. Find work done and PE gained.

Work = PE = 50 Γ— 9.8 Γ— 100
Work = 49000 J

Q61. A 150 kg box has PE = 7350 J. Find height.

7350 = 150 Γ— 9.8 Γ— h
h = 5 m

Q62. A 2 kg body is thrown up at 20 m/s. Find PE after 2 s.

Height = ut βˆ’ Β½gtΒ² = 40 βˆ’ 20 = 20 m
PE = 2 Γ— 10 Γ— 20 = 400 J

Q63. Find work done by 1 N force moving a body 1 m.

W = F Γ— s = 1 J

Q64. A force of 2.5Γ—10¹⁰ N moves a car for 2 minutes at 5 m/s. Find work done.

Distance = 5 Γ— 120 = 600 m
Work = F Γ— s = 1.5 Γ— 10ΒΉΒ³ J

Q65. A man does 2500 J work climbing a 5 m tree. Find mass.

2500 = m Γ— 10 Γ— 5
m = 50 kg

Q66. Work done is 24.2 J over 20 cm. Find force.

s = 0.2 m
F = W/s = 24.2 / 0.2
F = 121 N

Work Energy Power Important Numericals With Solution

Power – Class 9 (Solved Numericals)

70. A machine does 192 J of work in 24 Sec. What is the power of the machine?

Power = Work / Time = 192 / 24 = 8 W

71. A boy weighting 50 kg runs up a hill rising himself vertically 10 m in 20 Sec. Calculate power. given g = 9.8 m/sΒ²

Work = mgh = 50 Γ— 9.8 Γ— 10 = 4900 J
Power = 4900 / 20 = 245 W

72. A rickshaw puller pulls the rickshaw by applying a force of 100 N. If the rickshaw moves with constant velocity of 36 km h⁻¹. Find the power of rickshaw puller.

36 km/h = 10 m/s
Power = F Γ— v = 100 Γ— 10 = 1000 W

73. A athlete weighing 60 kg runs up a staircase having 10 steps each of 1 m in 30 sec. Calculate power (g = 9.8 m s⁻¹)

Height = 10 m
Work = 60 Γ— 9.8 Γ— 10 = 5880 J
Power = 5880 / 30 = 196 W

74. What is the power of a pump which takes 10 seconds to lift 100 kg of water to a water tank situated at a height of 20 m? (g = 10 m s⁻²)

Work = 100 Γ— 10 Γ— 20 = 20000 J
Power = 20000 / 10 = 2000 W = 2 kW

75. The heart does 1.5 J of work in each heartbeat. How many times per minute does it beat if its power is 2 watt?

Work per minute = 2 Γ— 60 = 120 J
Beats = 120 / 1.5 = 80 times

76. An electric bulb consumes 7.2 kJ of electrical energy in 2 minutes. What is the power of the electric bulb?

Energy = 7200 J, Time = 120 s
Power = 7200 / 120 = 60 W

77. When loading a truck, a man lifts boxes of 100 N each through a height of 1.5 m.

(a) Work = 100 Γ— 1.5 = 150 J
(b) Energy transferred = 150 J
(c) 4 boxes/min β†’ Energy = 600 J/min = 10 J/s
Power = 10 W

78. A man whose mass is 50 kg climbs up 30 steps of a stair in 30 s. If each step is 20 cm high, calculate the power used in climbing stairs

Height = 30 Γ— 0.2 = 6 m
Work = 50 Γ— 10 Γ— 6 = 3000 J
Power = 3000 / 30 = 100 W

79. A man drops a stone of mass 2 kg from the top of a building of height 15 m when it reaches the ground, find its kinetic energy. How?

KE = PE = mgh = 2 Γ— 10 Γ— 15 = 300 J

80. Two girls, each of weight 400 N climb up a rope through a height of 8 m. Girl A takes 20 s while B takes 50 s. What is the power expended by each girl?

Work = 400 Γ— 8 = 3200 J
Girl A: 3200 / 20 = 160 W
Girl B: 3200 / 50 = 64 W

81. A boy of mass 50 kg runs up a staircase of 45 steps in 9 s. If the height of each step is 15 cm, find his power. Take g = 10 m s⁻².

Height = 45 Γ— 0.15 = 6.75 m
Work = 50 Γ— 10 Γ— 6.75 = 3375 J
Power = 3375 / 9 = 375 W
Work energy powers formula sheet

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top