50 Physics Objective Questions on Work, Energy and Power (With Answers & Solutions)
50 Physics Objective Questions on Work, Energy and Power (With Answers & Solutions)
This comprehensive set of 50 multiple-choice questions covers the core topics of Work, Energy, and Power as taught in senior secondary physics. It is designed to help students preparing for JAMB, WAEC, NABTEB, and other examinations, as well as educators looking for ready-to-use practice resources. The questions test understanding of definitions, conditions for work, work done in gravitational fields, energy forms and conversions, kinetic and potential energy, conservation of mechanical energy, power calculations, and efficiency. Detailed solutions are provided for calculation-based questions to aid learning.
50 Physics Objective Questions on Work, Energy and Power
Questions
- Work is said to be done when a force applied to a bodyA. changes the shape of the bodyB. moves the body a distance in the direction of the forceC. is balanced by an equal and opposite forceD. acts on the body without causing motion
- Which of the following is a condition for work to be done?A. The force must be applied at an angle.B. The distance moved must be perpendicular to the force.C. The force must cause the body to move through a distance.D. The body must be at rest.
- A girl standing with a bucket of water on her head does no work becauseA. no force is appliedB. the bucket is too heavyC. there is no displacement in the direction of the forceD. the force is perpendicular to gravity
- The SI unit of work is theA. Newton (N)B. Joule (J)C. Watt (W)D. Pascal (Pa)
- A body of mass m moves in a circle of radius r with constant speed v. The work done by the centripetal force on the body isA. FvB. ZeroC. FmrD. Fr/m
- A load is pulled 5m along a horizontal floor by a constant force of 20N acting at 30° to the floor. Calculate the work done by the force.A. 10.0 JB. 17.3 JC. 50.0 JD. 86.6 J
- Work done in lifting a body of mass m through a height h in a gravitational field is given byA. maB. mghC. ½mv²D. Fs
- A load of mass 80kg fell down from the floor of a lorry 1.3m high. Taking g = 10ms⁻², the work done by gravity isA. 104 JB. 1040 JC. 80 JD. 800 J
- When a body falls freely from a height, the work done by gravity on it isA. mghB. ½mv²C. zeroD. mgh/2
- A man of mass 70kg climbs a vertical ladder of 20 steps each 0.25m high. The work done against gravity is (g = 10ms⁻²)A. 1750 JB. 3500 JC. 14000 JD. 35000 J
- For a body falling down an inclined plane of length l and inclination θ, the work done by gravity isA. mglB. mgl sinθC. mgl cosθD. mgh sinθ
- The area under a force–distance graph representsA. powerB. energyC. work doneD. efficiency
- In a force–extension graph for a spring obeying Hooke’s law, the work done in stretching the spring is given by the area of aA. rectangleB. triangleC. trapeziumD. circle
- A graph of force against distance for a constant force shows a rectangle of height 10N and width 4m. The work done isA. 40 JB. 14 JC. 2.5 JD. 0.4 J
- For a spring stretched by a force that increases uniformly from 0 to F, producing extension e, the work done isA. FeB. ½FeC. F/eD. 2Fe
- Which of the following is a renewable source of energy?A. CoalB. PetroleumC. NuclearD. Solar
- Energy is defined as theA. force applied to move a bodyB. capacity to do workC. rate of doing workD. product of force and distance
- The law of conservation of energy states that in an isolated system, total energyA. increases with timeB. decreases with timeC. remains constant, though may change formD. is always kinetic
- Which device converts mechanical energy to electrical energy?A. Electric motorB. GeneratorC. ThermocoupleD. Battery
- A steam engine primarily convertsA. chemical → heat → mechanical energyB. mechanical → electrical → heat energyC. heat → chemical → electrical energyD. electrical → heat → mechanical energy
- Gravitational potential energy of a body depends onA. mass and velocity onlyB. mass, height and acceleration due to gravityC. height and velocity onlyD. force and distance only
- An object at rest is said to possessA. kinetic energyB. potential energyC. electrical energyD. sound energy
- Kinetic energy of a body of mass m moving with velocity v isA. mvB. ½mvC. ½mv²D. mv²
- An object of mass 0.5kg has kinetic energy of 25J. Its speed isA. 50.0 ms⁻¹B. 25.0 ms⁻¹C. 10.0 ms⁻¹D. 5.0 ms⁻¹
- A body of mass 2kg is at a height of 5m above ground. Its potential energy is (g = 10ms⁻²)A. 10 JB. 50 JC. 100 JD. 200 J
- A body of mass 0.25kg moves at a height h with speed 4ms⁻¹. If its total mechanical energy is 12J, the value of h is (g = 10ms⁻²)A. 0.8 mB. 4.0 mC. 4.8 mD. 5.6 m
- A stone of mass 300g is released from rest from a height of 100m. The kinetic energy gained when it is a quarter way down is (g = 10ms⁻²)A. 750.0 JB. 225.0 JC. 75.0 JD. 22.5 J
- A body of mass 1000kg is released from a height of 10m. Its kinetic energy just before striking the ground is (g = 10ms⁻²)A. 10 JB. 10³ JC. 10⁴ JD. 10⁵ J
- For a swinging pendulum, at the lowest point of the swingA. kinetic energy is zero, potential energy maximumB. potential energy is zero, kinetic energy maximumC. kinetic energy equals potential energyD. total energy is zero
- A pendulum bob of mass 0.2kg is released from a height where the bob is 5cm above its lowest point. The maximum speed of the bob is (g = 10ms⁻²)A. 0.5 ms⁻¹B. 1.0 ms⁻¹C. 2.0 ms⁻¹D. 10.0 ms⁻¹
- Power is defined as theA. product of force and distanceB. rate of doing workC. ability to do workD. energy stored in a body
- The SI unit of power is theA. Joule (J)B. Watt (W)C. Newton (N)D. Pascal (Pa)
- A force of 20N pulls a body through 500m. If the power developed is 0.4 kW, the time taken isA. 250.0 sB. 25.0 sC. 2.5 sD. 0.5 s
- A car moving at a uniform velocity of 30ms⁻¹ overcomes a constant frictional force of 600N. The power of the engine isA. 18 kWB. 20 kWC. 180 kWD. 200 kW
- A man of mass 75kg ascends a vertical height of 44m while expending energy at 200 Js⁻¹. The time taken is (g = 10ms⁻²)A. 33.0 sB. 117.3 sC. 165.0 sD. 340.9 s
- A girl of mass 20kg climbs 25 steps each 15cm high in 10 seconds. The power expended is (g = 10ms⁻²)A. 7.5 WB. 75 WC. 750 WD. 7.5 kW
- A pump raises water from a depth of 20m to fill a reservoir of volume 1800m³ in 5 hours. If density of water is 1000 kgm⁻³ and g = 10ms⁻², the power of the pump isA. 20 WB. 200 WC. 2 kWD. 20 kW
- The expression for power in terms of force and velocity isA. P = F/sB. P = FvC. P = F/vD. P = v/F
- Efficiency of a machine is the ratio ofA. work output to work input, expressed as a percentageB. power input to power outputC. work input to work outputD. friction to useful work
- A machine lifts a load of 800N through 5m when 5000J of work is done on it. Its efficiency isA. 100%B. 90%C. 80%D. 75%
- A steam engine of efficiency 70% burns 20g of coal to produce 10kJ of energy. If it burns 200g of coal per second, its output power isA. 70 kWB. 80 kWC. 100 kWD. 200 kW
- Which of the following is a non-renewable energy source?A. WindB. SolarC. CoalD. Tidal
- When a man pushes a wall but the wall does not move, the work done isA. equal to the force appliedB. zeroC. equal to the man’s energyD. negative
- In a force–distance graph for a constant force, the work done is represented by theA. slope of the graphB. intercept on the force axisC. area under the graphD. length of the line
- A ball of mass 100g falls from 5m and rebounds to 3m. The energy lost is (g = 10ms⁻²)A. 2 JB. 20 JC. 100 JD. 2000 J
- Which of the following correctly states the principle of conservation of mechanical energy?A. The sum of kinetic and potential energy remains constant in an isolated system.B. Energy can be created but not destroyed.C. Potential energy always increases.D. Kinetic energy is always converted to heat.
- A 2kg object falls freely from rest. After falling 5m, its kinetic energy is (g = 10ms⁻²)A. 10 JB. 50 JC. 100 JD. 200 J
- A force of 50N pushes a box 8m across a floor in 4 seconds. The power developed isA. 100 WB. 200 WC. 400 WD. 1600 W
- An electric motor of efficiency 80% is used to lift a 20kg load through 6m. If g = 10ms⁻², the useful work done isA. 1200 JB. 960 JC. 1500 JD. 120 J
- A machine has an efficiency of 60%. If the power output is 300W, the power input isA. 180 WB. 500 WC. 360 WD. 3000 W
Answer Key
- B
- C
- C
- B
- B
- D
- B
- B
- A
- B
- B
- C
- B
- A
- B
- D
- B
- C
- B
- A
- B
- B
- C
- C
- C
- B
- C
- D
- B
- B
- B
- B
- B
- A
- C
- B
- D
- B
- A
- C
- A
- C
- B
- C
- A
- A
- C
- A
- A
- B
Solutions to Selected Questions
Q6: Work done = Fx × distance = (20 cos30°) × 5 = 20 × 0.8660 × 5 = 86.6 J. Answer D.
Q8: Work done by gravity = mgh = 80 × 10 × 1.3 = 1040 J. Answer B.
Q10: Total height = 20 × 0.25 = 5 m. Work = mgh = 70 × 10 × 5 = 3500 J. Answer B.
Q14: Work done = area of rectangle = 10 N × 4 m = 40 J. Answer A.
Q24: K.E = ½mv² → 25 = ½ × 0.5 × v² → v² = 100 → v = 10 ms⁻¹. Answer C.
Q26: M.E = mgh + ½mv² = 12 → 0.25×10×h + ½×0.25×4² = 12 → 2.5h + 2 = 12 → h = 4.0 m. Answer B.
Q27: m = 0.3 kg. Loss in P.E = mg(h/4) = 0.3×10×25 = 75 J = gain in K.E. Answer C.
Q28: P.E at top = K.E at ground = mgh = 1000×10×10 = 100,000 J = 10⁵ J. Answer D.
Q30: By conservation: mgh = ½mv² → v = √(2gh) = √(2×10×0.05) = √1 = 1 ms⁻¹. Answer B.
Q33: Power P = Fs/t → t = Fs/P = (20×500)/400 = 25 s. Answer B.
Q34: Power = Fv = 600 × 30 = 18,000 W = 18 kW. Answer A.
Q35: P = mgh/t → 200 = (75×10×44)/t → t = 33000/200 = 165 s. Answer C.
Q36: Total height = 25×0.15 = 3.75 m. P = mgh/t = (20×10×3.75)/10 = 75 W. Answer B.
Q37: m = ρV = 1000×1800 = 1.8×10⁶ kg. t = 5×3600 = 18000 s. P = mgh/t = (1.8×10⁶×10×20)/18000 = 20,000 W = 20 kW. Answer D.
Q40: Work output = load × height = 800×5 = 4000 J. Efficiency = (4000/5000)×100% = 80%. Answer C.
Q41: 20g coal → 10 kJ. Burning 200g/s gives (200/20)×10 = 100 kJ/s = 100,000 W input. Output = 70% of input = 0.7×100,000 = 70,000 W = 70 kW. Answer A.
Q45: Energy lost = mg(h₁ – h₂) = 0.1×10×(5-3) = 2 J. Answer A.
Q47: Loss in P.E = mgh = 2×10×5 = 100 J = gain in K.E. Answer C.
Q48: Work = Fs = 50×8 = 400 J. Power = work/time = 400/4 = 100 W. Answer A.
Q49: Useful work output = mgh = 20×10×6 = 1200 J. Note: Efficiency is not needed here; the question asks for useful work done (output). Answer A.
Q50: Efficiency = (Power output / Power input)×100% → 60 = (300/P_in)×100 → P_in = 300/0.6 = 500 W. Answer B.


