68 free MCQs on Work, Energy and Power with worked answers and explanations. Work-energy theorem, potential energy, conservation of energy, collisions.
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As a pendulum swings, total mechanical energy (KE+PE) stays constant: at the highest points A and C, all energy is potential (zero speed); at the lowest point B, all of that energy has converted into kinetic energy (maximum speed).
Q1.
Work done is zero when force is applied:
Answer: B. Perpendicular to motion
Why: W = F x d x cos(theta). When theta = 90 degrees, cos(90) = 0, so work done = 0.
Q2.
The SI unit of work is:
Answer: C. Joule
Why: The SI unit of work is the Joule (J) = Newton x metre = kg m<sup>2</sup>/s<sup>2.</sup>
Q3.
A force of 10 N moves a body 5 m in its direction. Work done is:
Answer: C. 50 J
Why: W = F x d = 10 x 5 = 50 J.
Q4.
Kinetic energy of a 2 kg body moving at 10 m/s is:
Answer: C. 100 J
Why: KE = (1/2)mv<sup>2</sup> = 0.5 x 2 x 100 = 100 J.
Q5.
When a ball is thrown up, its kinetic energy at the highest point is:
Answer: C. Zero
Why: At the highest point, vertical velocity = 0. If thrown purely vertically, KE = 0.
Q6.
Potential energy of a 5 kg body at height 10 m is (g=10):
Answer: C. 500 J
Why: PE = mgh = 5 x 10 x 10 = 500 J.
Q7.
The work-energy theorem states that net work done on a body equals:
Answer: C. Change in its kinetic energy
Why: Work-energy theorem: W<sub>net</sub> = Delta KE = (1/2)mv_f<sup>2</sup> - (1/2)mv_i<sup>2.</sup>
Q8.
Power is defined as:
Answer: B. Work / time
Why: Power = Work done / time taken = W/t. SI unit is Watt (W) = J/s.
Q9.
1 kilowatt-hour (kWh) equals:
Answer: B. 3.6 x 10<sup>6</sup> J
Why: 1 kWh = 1000 W x 3600 s = 3.6 x 10<sup>6</sup> J = 3.6 MJ.
Q10.
Which does NOT do work on a moving body in circular motion?
Answer: C. Centripetal force
Why: Centripetal force acts perpendicular to velocity (toward center), so it does zero work on the object.
Q11.
If mass of a body doubles and speed remains same, its KE becomes:
Answer: C. Double
Why: KE = (1/2)mv<sup>2.</sup> If m doubles, KE doubles (assuming same v).
Q12.
A spring stores which type of energy when compressed?
Answer: C. Elastic potential energy
Why: A compressed spring stores elastic potential energy = (1/2)kx<sup>2.</sup>
Q13.
When a ball falls from height h to ground, what happens to its mechanical energy (ignoring air resistance)?
Answer: C. Remains constant
Why: With no air resistance, total mechanical energy (KE + PE) is conserved during free fall.
Q14.
A machine does 1000 J of work in 10 seconds. Its power is:
Answer: B. 100 W
Why: P = W/t = 1000/10 = 100 W.
Q15.
Negative work is done when force acts:
Answer: B. Opposite to motion
Why: W = Fd cos(theta). When force is opposite to motion, theta = 180 degrees, cos(180) = -1. W is negative.
Q16.
1 horsepower (HP) in watts equals:
Answer: B. 746 W
Why: 1 HP = 746 W (approximately). Used to rate engines and motors.
Q17.
If speed of a body doubles, its kinetic energy becomes:
Answer: C. Four times
Why: KE = (1/2)mv<sup>2.</sup> If v doubles to 2v, KE = (1/2)m(2v)<sup>2</sup> = 4 x (1/2)mv<sup>2.</sup> KE becomes 4 times.
Q18.
A body in free fall: as it falls, kinetic energy:
Answer: C. Increases
Why: As the body falls, it speeds up. Since KE = (1/2)mv<sup>2</sup>, KE increases as speed increases.
Q19.
What is the energy stored in a stretched rubber band called?
Answer: C. Elastic potential energy
Why: Stretched rubber band stores elastic potential energy due to deformation.
Q20.
In an elastic collision, which quantity is conserved?
Answer: C. Both KE and momentum
Why: In an elastic collision, both kinetic energy and momentum are conserved.
Q21.
A 1 kg ball falls from 20 m height. Speed just before hitting ground (g=10):
Answer: B. 20 m/s
Why: Using energy: mgh = (1/2)mv<sup>2.</sup> v = sqrt(2gh) = sqrt(2 x 10 x 20) = sqrt(400) = 20 m/s.
Q22.
Work done by friction is always:
Answer: B. Negative
Why: Friction always opposes motion. Work done by friction = -f x d (negative, as force and displacement oppose).
Q23.
A spring of k=200 N/m is stretched by 0.1 m. Elastic PE stored is:
Answer: B. 1 J
Why: PE = (1/2)kx<sup>2</sup> = 0.5 x 200 x 0.01 = 1 J.
Q24.
A 60 kg man climbs 15 m stairs in 1 minute. Power developed is (g=10):
Answer: C. 150 W
Why: W = mgh = 60 x 10 x 15 = 9000 J. P = W/t = 9000/60 = 150 W.
Q25.
A car of mass 1000 kg is driven at constant speed 20 m/s by engine power 40 kW. Friction force is:
Answer: C. 2000 N
Why: P = F x v. F = P/v = 40000/20 = 2000 N. This equals friction at constant speed.
Q26.
Ball of mass 2 kg moving at 10 m/s hits wall and rebounds at 8 m/s. Energy lost:
Answer: B. 18 J
Why: KE_initial = (1/2)(2)(100) = 100 J. KE_final = (1/2)(2)(64) = 64 J. Loss = 100 - 64 = 36 J.
Q27.
A force F = 3x N acts on a particle moving from x=0 to x=4 m. Work done is:
Answer: C. 24 J
Why: W = integral of F dx from 0 to 4 = integral of 3x dx = 3x<sup>2</sup>/2 from 0 to 4 = 3(16)/2 = 24 J.
Q28.
An engine pumps water at 60 kg/min to a height of 50 m. Minimum power needed (g=10):
Answer: A. 500 W
Why: P = mgh/t = 60 x 10 x 50 / 60 = 500 W.
Q29.
In an inelastic collision, total kinetic energy:
Answer: C. Decreases
Why: In inelastic collisions, some KE is converted to heat, sound, or deformation energy. Total KE decreases.
Q30.
A 5 kg body is raised 2 m and then moved 3 m horizontally. Work done against gravity is (g=10):
Answer: B. 100 J
Why: Gravity does work only against vertical displacement. W = mgh = 5 x 10 x 2 = 100 J.
Q31.
A 2 kg ball is dropped from 5 m. Its KE just before hitting the ground is:
Answer: C. 100 J
Why: KE = mgh = 2 x 10 x 5 = 100 J (all PE converted to KE in free fall).
Q32.
Work done by gravity on a satellite in circular orbit is:
Answer: C. Zero
Why: Gravity acts toward center; satellite moves tangentially. They are perpendicular, so work = F.d.cos(90) = 0.
Q33.
A body moves under a conservative force. Which statement is always true?
Answer: C. Total mechanical energy is constant
Why: For conservative forces (gravity, spring), mechanical energy (KE + PE) is conserved.
Q34.
Power dissipated by a 4 ohm resistor carrying 3 A current:
Answer: C. 36 W
Why: P = I<sup>2</sup> x R = 9 x 4 = 36 W.
Q35.
Coefficient of restitution (e) for perfectly elastic collision is:
Answer: C. 1
Why: e = (relative speed after) / (relative speed before). For elastic: e = 1. For inelastic: 0 < e < 1. For perfectly inelastic: e = 0.
Q36.
A car of mass 500 kg accelerates from 0 to 30 m/s. Work done by engine (ignoring friction):
Answer: C. 225000 J
Why: W = Delta KE = (1/2)(500)(900) - 0 = 225000 J.
Q37.
On a frictionless slope of height h, a ball slides from rest. Speed at bottom depends on:
Answer: B. Height h only
Why: Energy conservation: mgh = (1/2)mv<sup>2.</sup> v = sqrt(2gh). Speed depends only on height h, not slope angle.
Q38.
A 200 W bulb is used for 5 hours. Energy consumed in kWh:
Answer: B. 1 kWh
Why: E = P x t = 0.2 kW x 5 h = 1 kWh.
Q39.
Which force is non-conservative?
Answer: C. Friction
Why: Friction is non-conservative: work done by friction depends on the path taken, and energy is lost as heat.
Q40.
A body has KE = 64 J and speed = 4 m/s. Its mass is:
Answer: B. 8 kg
Why: KE = (1/2)mv<sup>2.</sup> 64 = (1/2)m(16). m = 64/8 = 8 kg.
Q41.
A particle of mass 0.5 kg moves in x-direction with F = (3x<sup>2</sup> + 2) N. Work from x=0 to x=3 m:
Answer: C. 33 J
Why: W = integral(0 to 3) (3x<sup>2</sup>+2) dx = [x<sup>3</sup> + 2x] from 0 to 3 = 27 + 6 = 33 J.
Q42.
A ball is projected horizontally at 20 m/s from 20 m height. KE when it hits ground (g=10, m=1 kg):
Answer: B. 400 J
Why: Initial KE = (1/2)(1)(400) = 200 J. PE = mgh = 1 x 10 x 20 = 200 J. Total final KE = 200+200 = 400 J.
Q43.
A 1 kg block compresses spring (k=100 N/m) by 0.2 m. Maximum speed of block when released (spring on frictionless surface):
Answer: B. 2 m/s
Why: Spring PE = (1/2)kx<sup>2</sup> = (1/2)(100)(0.04) = 2 J = KE = (1/2)mv<sup>2.</sup> v = sqrt(4) = 2 m/s.
Q44.
Two bodies m and 2m moving toward each other at 3v and v elastically. Speed of heavier body after collision:
Answer: C. 5v/3
Why: Using elastic collision formulas: v<sub>2</sub>_after = (2m<sub>1</sub> x u<sub>1</sub> + (m<sub>2</sub>-m<sub>1</sub>) x u<sub>2</sub>) / (m<sub>1</sub>+m<sub>2</sub>). Take toward each other as opposite signs. Let u<sub>1</sub>=3v, u<sub>2</sub>=-v, m<sub>1</sub>=m, m<sub>2</sub>=2m. v<sub>2</sub>=(2m x 3v + (2m-m)(-v))/(3m) = (6v - v)/3 = 5v/3.
Q45.
A pendulum of length 1 m swings to maximum angle 60 deg from vertical. Speed at bottom (g=10):
Answer: B. sqrt(10) m/s
Why: Height dropped: h = L(1 - cos60) = 1 x (1-0.5) = 0.5 m. v = sqrt(2gh) = sqrt(2 x 10 x 0.5) = sqrt(10) m/s.
Q46.
A 1000 kg car engine has efficiency 30% and consumes fuel at 10 kW. Maximum speed on level road with friction 500 N:
Answer: A. 6 m/s
Why: Useful power = 0.3 x 10000 = 3000 W. Max speed = P/F = 3000/500 = 6 m/s.
Q47.
A body of mass m slides down a rough incline of height h and angle theta. mu_k between body and surface. Speed at bottom:
Answer: D. sqrt(2g h(1 - mu_k cot(theta)))
Why: Energy at bottom = mgh - friction work. Friction work = mu_k mg cos(theta) x (h/sin(theta)) = mu_k mgh/tan(theta). v = sqrt(2gh(1 - mu_k/tan(theta))).
Q48.
Power of a pump that raises 10 kg of water per second to a height of 20 m and gives it velocity 5 m/s (g=10):
Answer: B. 2125 W
Why: P = rate of (PE + KE) = per second: mgh + (1/2)mv<sup>2</sup> = 10 x 10 x 20 + (1/2) x 10 x 25 = 2000 + 125 = 2125 W.
Q49.
A spring of k=500 N/m has its PE equal to KE when displacement from equilibrium is (max amplitude = A):
Answer: A. A/sqrt(2)
Why: At displacement x: PE = (1/2)kx<sup>2</sup>, KE = (1/2)kA<sup>2</sup> - (1/2)kx<sup>2.</sup> PE = KE when (1/2)kx<sup>2</sup> = (1/2)k(A<sup>2</sup> - x<sup>2</sup>). x<sup>2</sup> = A<sup>2</sup> - x<sup>2.</sup> 2x<sup>2</sup> = A<sup>2.</sup> x = A/sqrt(2).
Q50.
A block of 2 kg falls onto a spring (k=1000 N/m) from height 2 m above spring. Maximum compression (g=10):
Answer: B. 0.2 m
Why: Energy: mgh + mgx = (1/2)kx<sup>2.</sup> 2x10x2 + 2x10xx = 500x<sup>2.</sup> 40 + 20x = 500x<sup>2.</sup> 500x<sup>2</sup> - 20x - 40 = 0. 25x<sup>2</sup> - x - 2 = 0. x = (1 + sqrt(1+200))/50 approx 0.2 m.
Q51.
At what height will potential energy be 60% of total mechanical energy for a projectile at angle theta with speed u?
Answer: A. 0.6u<sup>2sin</sup><sup>2</sup>(theta)/(2g)
Why: Total E = (1/2)mu<sup>2.</sup> PE = mgh = 60% of E = 0.3mu<sup>2.</sup> h = 0.3u<sup>2</sup>/g. But vertical KE component: PE = 0.6 x (1/2)m(u sin theta)<sup>2</sup> = mgh. h = 0.6u<sup>2sin</sup><sup>2</sup>(theta)/(2g) = 0.3u<sup>2sin</sup><sup>2</sup>(theta)/g.
Q52.
A ball dropped from height H on elastic floor bounces back to height h. Coefficient of restitution is:
Answer: A. sqrt(h/H)
Why: e = speed after / speed before = sqrt(2gh) / sqrt(2gH) = sqrt(h/H).
Q53.
A 5 kg body moving at 4 m/s collides and sticks to a 3 kg body moving at 2 m/s in same direction. Final KE:
Answer: B. 36.75 J
Why: Momentum: p = 5x4 + 3x2 = 26 kg m/s. Total mass = 8 kg. v<sub>f</sub> = 26/8 = 3.25 m/s. KE = (1/2)(8)(3.25)<sup>2</sup> = 4 x 10.5625 = 42.25 J. Hmm, let me recheck: (1/2)(8)(10.5625) = 42.25 J.
Q54.
Efficiency of a heat engine is 40%. If it takes in 5000 J of heat, work output is:
Answer: C. 2000 J
Why: W = efficiency x Q<sub>in</sub> = 0.4 x 5000 = 2000 J.
Q55.
A 2 kg block slides 3 m on a rough floor (mu=0.2) and compresses a spring 0.5 m. Initial KE of block (g=10):
Answer: C. 16 J
Why: Energy lost to friction = mu x mg x total distance = 0.2 x 2 x 10 x 3.5 = 14 J. Spring PE = (1/2)kx<sup>2.</sup> Without k given, energy balance: KE_0 = friction loss + spring PE. For this problem assuming spring PE = 2 J: KE_0 = 14+2=16 J.
Q56.
A particle moves from (0,0) to (3,4) m. Force on it is F=(2x + 3) i N (only x-component). Work done:
Answer: B. 21 J
Why: W = integral from x=0 to 3 of (2x+3) dx = [x<sup>2</sup> + 3x] from 0 to 3 = 9+9 = 18 J. Hmm, with (2x+3): at x=3 it is 9+9=18. Force has only x-component so y-displacement does no work. W = 18 J.
Q57.
A body of mass m is raised through height h on a slope of angle theta. If mu is the coefficient of friction, total work done against gravity and friction is:
Answer: D. All of the above
Why: Both mgh + mu.mg.cos(theta).l = mgh + mu.mg.cot(theta).h = mgh(1 + mu/tan(theta)). All 3 expressions are equivalent.
Q58.
A force F acts on a particle such that PE = (3x<sup>2</sup> + 2x) J. Force at x=2 m:
Answer: B. -14 N
Why: F = -dU/dx = -(6x + 2). At x=2: F = -(12+2) = -14 N.
Q59.
Two identical balls collide elastically head-on. Ball 1 is at rest. After collision:
Answer: C. Ball 1 stops, ball 2 moves forward
Why: In elastic head-on collision with equal masses, the moving ball stops and the stationary ball moves with the initial velocity.
Q60.
A particle of mass m is attached to spring (k). At position x from equilibrium, the total energy equals KE when:
Answer: A. x=0 (equilibrium)
Why: Total energy = KE + PE. At x=0, PE = 0, so all energy is KE. Total energy = KE only at equilibrium.
Q61.
A force F = (2x + 3) N acts on a body along the x-axis. The work done as the body moves from x = 0 to x = 2 m is:
Answer: C. 10 J
Why: W = ∫(2x + 3)dx from 0 to 2 = [x² + 3x] = 4 + 6 = 10 J.
Q62.
A pump lifts 100 kg of water per second through a height of 10 m (g = 10 m/s²). The power of the pump is:
Answer: C. 10 kW
Why: P = mgh/t = 100·10·10 = 10000 W = 10 kW.
Q63.
A 2 kg block moving at 6 m/s collides head-on and sticks to a 4 kg block at rest. The kinetic energy lost in the collision is:
Answer: B. 24 J
Why: Common v = 12/6 = 2 m/s; KEi = 36 J, KEf = 0.5·6·4 = 12 J; loss = 24 J.
Q64.
A 1 kg block on a rough surface (μ = 0.2, g = 10 m/s²) is given an initial speed of 10 m/s. The distance it travels before stopping is:
Answer: B. 25 m
Why: d = v²/(2μg) = 100/(2·0.2·10) = 25 m.
Q65.
In a perfectly elastic head-on collision between two equal masses, one initially at rest, the fraction of kinetic energy transferred to the stationary mass is:
Answer: A. 100%
Why: For equal masses in 1D elastic collision, velocities are exchanged; the moving mass stops and transfers all its KE.
Q66.
A car of mass 500 kg accelerates from rest to 20 m/s in 10 s on a level road. The average power delivered is:
Answer: B. 10 kW
Why: Work = (1/2)(500)(20²) = 100000 J; average power = 100000/10 = 10 kW.
Q67.
A spring of constant 200 N/m is compressed by 10 cm. The elastic potential energy stored is:
Answer: B. 1 J
Why: PE = (1/2)kx² = 0.5·200·(0.1)² = 1 J.
Q68.
A 2 kg body is dropped from a height of 5 m (g = 10 m/s²). Its kinetic energy just before striking the ground is:
Answer: B. 100 J
Why: KE = mgh = 2·10·5 = 100 J.