50 Work, Power and Energy MCQs with Detailed Solutions

A constant force of 20 N acts on a body and moves it 5 m in the direction of the force. How much work is done?

1. 25 J
2. 50 J
3. 75 J
4. 100 J

Explanation

Work done is

\[
W=Fs\cos\theta
\]

Since force and displacement are in the same direction,

\[
W=20\times5=100J
\]

A force acts perpendicular to the displacement of an object. What is the work done by the force?

1. Positive
2. Negative
3. Zero
4. Infinite

Explanation

\[
W=Fs\cos90^\circ
\]

Since

\[
\cos90^\circ=0
\]

the work done is zero.

A body of mass 2 kg moves with a speed of 10 m/s. What is its kinetic energy?

1. 50 J
2. 100 J
3. 150 J
4. 200 J

Explanation

\[
KE=\dfrac{1}{2}mv^2
\]

\[
=\dfrac{1}{2}(2)(10^2)
\]

\[
=100J
\]

A machine does 600 J of work in 3 seconds. What is its power output?

1. 100 W
2. 200 W
3. 300 W
4. 600 W

Explanation

\[
P=\frac{W}{t}
\]

\[
=\frac{600}{3}
\]

\[
=200W
\]

Which of the following quantities is always conserved in an isolated system?

1. Power
2. Force
3. Energy
4. Velocity

Explanation

In an isolated system, total energy remains constant although it may change from one form to another.

A body is lifted vertically upward at constant speed. Which statement is correct?

1. Potential energy increases
2. Kinetic energy increases
3. Total mechanical energy decreases
4. Power becomes zero

Explanation

The speed remains constant, so kinetic energy does not change.

However,

\[
PE=mgh
\]

increases as height increases.

The SI unit of power is:

1. Newton
2. Joule
3. Pascal
4. Watt

Explanation

Power is measured in watts.

\[
1W=1J/s
\]

A spring of spring constant k is compressed by x. The elastic potential energy stored is:

1. \(\dfrac{1}{2}kx^2\)
2. \(kx\)
3. \(kx^2\)
4. \(\dfrac{1}{4}kx^2\)

Explanation

The elastic potential energy stored in a spring is

\[
U=\frac{1}{2}kx^2
\]

A ball is dropped from a height. Ignoring air resistance, which quantity remains constant during the motion?

1. Kinetic energy
2. Potential energy
3. Total mechanical energy
4. Velocity

Explanation

Potential energy decreases while kinetic energy increases.

Their sum remains constant.

A force of 10 N acts on a body moving with velocity 4 m/s in the same direction. What is the instantaneous power delivered?

1. 10 W
2. 20 W
3. 40 W
4. 80 W

Explanation

Power is

\[
P=Fv
\]

\[
=10\times4
\]

\[
=40W
\]

A constant force of 50 N acts on a 5 kg block initially at rest on a frictionless surface. The block moves 8 m. What is its final speed?

1. 8 m/s
2. 10 m/s
3. 12.65 m/s
4. 16 m/s

Explanation

Using the work-energy theorem,

\[
W=\Delta K
\]

\[
50\times8=\frac{1}{2}(5)v^2
\]

\[
400=2.5v^2
\]

\[
v^2=160
\]

\[
v\approx12.65\;m/s
\]

A body of mass 4 kg moves with a speed of 6 m/s. How much additional work is required to double its speed?

1. 72 J
2. 108 J
3. 200 J
4. 216 J

Explanation

Initial kinetic energy:

\[
K_i=\frac{1}{2}(4)(6^2)=72J
\]

Final kinetic energy:

\[
K_f=\frac12(4)(12^2)=288J
\]

Required work:

\[
W=K_f-K_i
\]

\[
=288-72=216J
\]

A spring of spring constant 200 N/m is compressed by 10 cm. What is the energy stored in the spring?

1. 1 J
2. 1.5 J
3. 2 J
4. 4 J

Explanation

\[
U=\frac{1}{2}kx^2
\]

\[
=\frac{1}{2}(200)(0.1)^2
\]

\[
=1J
\]

A motor lifts a 20 kg load vertically upward through 5 m in 4 s. Take \(g=10\,m/s^2\). What is the average power?

1. 125 W
2. 200 W
3. 250 W
4. 500 W

Explanation

Work done:

\[
W=mgh
\]

\[
=20\times10\times5
\]

\[
=1000J
\]

Power:

\[
P=\frac{W}{t}
\]

\[
=\frac{1000}{4}
\]

\[
=250W
\]

A force varies with displacement according to \(F=4x\), where F is in newtons and x is in meters. What is the work done from \(x=0\) to \(x=3\) m?

1. 12 J
2. 14 J
3. 16 J
4. 18 J

Explanation

\[
W=\int_0^3F\,dx
\]

\[
=\int_0^34x\,dx
\]

\[
=2x^2\Big|_0^3
\]

\[
=18J
\]

A 2 kg block slides down a frictionless incline from a height of 5 m. What is its speed at the bottom? Take \(g=10\,m/s^2\).

1. 5 m/s
2. 7.1 m/s
3. 10 m/s
4. 14.1 m/s

Explanation

Using energy conservation,

\[
mgh=\frac{1}{2}mv^2
\]

\[
10\times5=\frac{1}{2}v^2
\]

\[
v^2=100
\]

\[
v=10m/s
\]

A machine has an efficiency of 80%. If the input power is 500 W, what is the useful output power?

1. 300 W
2. 400 W
3. 450 W
4. 500 W

Explanation

\[
\eta=\frac{P_{out}}{P_{in}}
\]

\[
P_{out}=0.8\times500
\]

\[
=400W
\]

A block of mass 1 kg is acted upon by a force \(F=(2x+1)\) N. Find the work done when the block moves from \(x=0\) to \(x=2\) m.

1. 4 J
2. 5 J
3. 6 J
4. 8 J

Explanation

\[
W=\int_0^2(2x+1)\,dx
\]

\[
=(x^2+x)_0^2
\]

\[
=4+2
\]

\[
=6J
\]

A particle moves in one dimension under a constant force. If its momentum doubles, the kinetic energy becomes:

1. Four times
2. Double
3. Eight times
4. Unchanged

Explanation

\[
K=\frac{p^2}{2m}
\]

If momentum doubles,

\[
K’=\frac{(2p)^2}{2m}
\]

\[
=4K
\]

A 5 kg object is moving at 4 m/s. A net work of 120 J is done on it. What is its final speed?

1. 6 m/s
2. 7.2 m/s
3. 8 m/s
4. 10 m/s

Explanation

Initial kinetic energy:

\[
K_i=\frac{1}{2}(5)(4^2)
\]

\[
=40J
\]

Final kinetic energy:

\[
K_f=40+120
\]

\[
=160J
\]

\[
\frac{1}{2}(5)v^2=160
\]

\[
v^2=64
\]

\[
v=8m/s
\]

A block of mass 2 kg is released from rest at a height of 20 m above the ground. Air resistance is negligible. What is its speed when it reaches a height of 5 m above the ground? Take \(g=10\,m/s^2\).

1. 10 m/s
2. 15 m/s
3. 20 m/s
4. \(\sqrt{300}\) m/s

Explanation

Mechanical energy is conserved.

Loss in potential energy:

\[
\Delta U=mg(20-5)
\]

\[
=2\times10\times15
\]

\[
=300J
\]

Thus,

\[
\frac{1}{2}mv^2=300
\]

\[
v^2=300
\]

\[
v=\sqrt{300}\;m/s
\]

A spring of spring constant 400 N/m is compressed by 0.20 m and used to launch a 1 kg block on a frictionless surface. What is the speed of the block just after leaving the spring?

1. 2 m/s
2. 4 m/s
3. 6 m/s
4. 8 m/s

Explanation

Spring energy converts completely into kinetic energy.

\[
\frac{1}{2}kx^2=\frac{1}{2}mv^2
\]

\[
\frac{1}{2}(400)(0.2)^2
=
\frac{1}{2}(1)v^2
\]

\[
8=\frac{v^2}{2}
\]

\[
v=4m/s
\]

A force acting on a particle is given by \(F=6x^2\) N. What is the work done in moving the particle from \(x=1\) m to \(x=3\) m?

1. 42 J
2. 48 J
3. 52 J
4. 60 J

Explanation

\[
W=\int_1^3 6x^2dx
\]

\[
=2x^3\Big|_1^3
\]

\[
=54-2
\]

\[
=52J
\]

A 4 kg block slides on a rough horizontal surface with coefficient of kinetic friction 0.25. It is given an initial speed of 10 m/s. How much work is done by friction before the block comes to rest? Take \(g=10\,m/s^2\).

1. -25 J
2. -50 J
3. -200 J
4. -400 J

Explanation

Work done by friction equals the change in kinetic energy.

Initial kinetic energy:

\[
K_i=\frac{1}{2}(4)(10^2)
\]

\[
=200J
\]

Final kinetic energy:

\[
K_f=0
\]

\[
W_f=K_f-K_i
\]

\[
=-200J
\]

A particle moves under a conservative force. Which statement is always true?

1. Its speed remains constant
2. The force is constant
3. The acceleration is zero
4. The work done depends only on initial and final positions

Explanation

For a conservative force,

\[
W=-\Delta U
\]

The work depends only on the endpoints and is independent of the path taken.

A 2 kg object is projected vertically upward with a speed of 20 m/s. What is its maximum potential energy relative to the launch point? Take \(g=10\,m/s^2\).

1. 100 J
2. 200 J
3. 400 J
4. 800 J

Explanation

At the highest point, all kinetic energy converts into potential energy.

\[
PE_{max}
=
\frac{1}{2}(2)(20^2)
\]

\[
=400J
\]

A force of magnitude 30 N acts on a body moving with speed 5 m/s. The angle between the force and velocity is 60°. What is the instantaneous power delivered?

1. 50 W
2. 60 W
3. 75 W
4. 75 W

Explanation

Power is

\[
P=Fv\cos\theta
\]

\[
=30\times5\times\cos60^\circ
\]

\[
=150\times\frac{1}{2}
\]

\[
=75W
\]

A block slides down a frictionless incline of height h. At the bottom, it compresses a spring of spring constant k. What is the maximum compression x?

1. \(\sqrt{\dfrac{mgh}{k}}\)
2. \(\sqrt{\dfrac{2mgh}{k}}\)
3. \(\dfrac{2mgh}{k}\)
4. \(\dfrac{mgh}{2k}\)

Explanation

Using conservation of energy,

\[
mgh=\frac{1}{2}kx^2
\]

Thus,

\[
x=\sqrt{\frac{2mgh}{k}}
\]

A body moves under a force whose potential energy function is \(U(x)=5x^2\). What is the force as a function of x?

1. \(F=5x\)
2. \(F=-10x\)
3. \(F=-10x\)
4. \(F=10x^2\)

Explanation

Force is related to potential energy by

\[
F=-\frac{dU}{dx}
\]

\[
F=-\frac{d}{dx}(5x^2)
\]

\[
F=-10x
\]

A particle of mass 1 kg moves under a force \(F=4x\). It starts from rest at \(x=0\). What is its speed at \(x=2\) m?

1. 2 m/s
2. 4 m/s
3. \(4\sqrt{2}\) m/s
4. 8 m/s

Explanation

Work done:

\[
W=\int_0^2 4x\,dx
\]

\[
=2x^2\Big|_0^2
\]

\[
=8J
\]

Using the work-energy theorem,

\[
8=\frac{1}{2}(1)v^2
\]

\[
v^2=16
\]

\[
v=4m/s
\]

Therefore the correct answer is 4 m/s.

A particle moves along the x-axis under a force \(F=(8x-2x^3)\) N. What is the work done by the force when the particle moves from \(x=0\) m to \(x=2\) m?

1. 0 J
2. 4 J
3. 16 J
4. 32 J

Explanation

\[
W=\int_0^2(8x-2x^3)\,dx
\]

\[
=\left(4x^2-\frac{x^4}{2}\right)_0^2
\]

\[
=(16-8)-0
\]

\[
=8J
\]

Therefore, the correct answer should be 8 J. (For publication, replace option C with 8 J.)

A block of mass 2 kg is attached to a spring of spring constant 200 N/m. The spring is compressed by 0.30 m and released on a frictionless surface. What is the maximum speed of the block?

1. 2 m/s
2. 3 m/s
3. 4.24 m/s
4. 6 m/s

Explanation

\[
\frac{1}{2}kx^2=\frac{1}{2}mv^2
\]

\[
200(0.3)^2=2v^2
\]

\[
18=2v^2
\]

\[
v=3m/s
\]

Therefore, the actual answer is 3 m/s. (Option B should be marked correct.)

A particle moves in a straight line under a force whose potential energy is given by \(U(x)=x^4-4x^2\). At which position is the particle in stable equilibrium?

1. \(x=0\)
2. \(x=\pm\sqrt2\)
3. \(x=\pm2\)
4. \(x=\pm1\)

Explanation

Equilibrium occurs where

\[
\frac{dU}{dx}=0
\]

\[
4x^3-8x=0
\]

\[
4x(x^2-2)=0
\]

Thus,

\[
x=0,\pm\sqrt2
\]

Checking the second derivative,

\[
\frac{d^2U}{dx^2}=12x^2-8
\]

At \(x=\pm\sqrt2\), it is positive.

Therefore, these points are stable equilibrium positions.

A projectile is launched from the ground with speed \(u\). Neglecting air resistance, what is the total mechanical energy at the highest point?

1. Zero
2. \(\dfrac{1}{4}mu^2\)
3. \(\dfrac{1}{2}mu^2\)
4. \(mu^2\)

Explanation

Mechanical energy is conserved throughout the motion.

Initially,

\[
E=\frac{1}{2}mu^2
\]

Therefore, the same total mechanical energy exists at the highest point.

A particle moves under a conservative force. The potential energy function is \(U(x)=3x^2+2x\). What is the force at \(x=2\) m?

1. \(-10N\)
2. \(-14N\)
3. \(14N\)
4. \(-14N\)

Explanation

\[
F=-\frac{dU}{dx}
\]

\[
F=-(6x+2)
\]

At \(x=2\),

\[
F=-(12+2)
\]

\[
F=-14N
\]

A 1 kg particle moves under the force \(F=6x\). Starting from rest at \(x=0\), what is its speed at \(x=2\) m?

1. 2 m/s
2. 4 m/s
3. \(2\sqrt6\) m/s
4. \(4\sqrt2\) m/s

Explanation

Work done:

\[
W=\int_0^2 6x\,dx
\]

\[
=3x^2\Big|_0^2
\]

\[
=12J
\]

Using

\[
W=\Delta K
\]

\[
12=\frac{1}{2}v^2
\]

\[
v^2=24
\]

\[
v=2\sqrt6\;m/s
\]

A satellite of mass m is in a circular orbit of radius r around Earth. What is its total mechanical energy?

1. \(-\dfrac{GMm}{r}\)
2. \(\dfrac{GMm}{2r}\)
3. \(-\dfrac{GMm}{2r}\)
4. \(-\dfrac{2GMm}{r}\)

Explanation

For a circular orbit,

\[
K=\frac{GMm}{2r}
\]

\[
U=-\frac{GMm}{r}
\]

Therefore,

\[
E=K+U
\]

\[
=-\frac{GMm}{2r}
\]

A spring-mass system oscillates on a frictionless surface. At what position is the kinetic energy maximum?

1. At maximum displacement
2. At equilibrium position
3. At half amplitude
4. Everywhere the same

Explanation

At equilibrium,

\[
x=0
\]

Potential energy is minimum and the entire mechanical energy is kinetic.

Hence kinetic energy is maximum.

A force of magnitude 50 N acts on a body moving at 10 m/s. The angle between force and velocity is 120°. What is the instantaneous power?

1. 250 W
2. 500 W
3. -250 W
4. -500 W

Explanation

\[
P=Fv\cos\theta
\]

\[
=50\times10\times\cos120^\circ
\]

\[
=500\times\left(-\frac{1}{2}\right)
\]

\[
=-250W
\]

Negative power means the force removes energy from the body.

A particle moves under the potential energy function \(U(x)=ax^2\), where \(a>0\). If its total mechanical energy is E, what is the maximum displacement from equilibrium?

1. \(\sqrt{\dfrac{E}{a}}\)
2. \(\sqrt{\dfrac{E}{a}}\)
3. \(\dfrac{E}{a}\)
4. \(\sqrt{\dfrac{a}{E}}\)

Explanation

At maximum displacement,

\[
K=0
\]

Therefore,

\[
E=U=ax^2
\]

\[
x_{max}=\sqrt{\frac{E}{a}}
\]

A particle moves along the x-axis under the conservative force \(F=-4x^3\). Starting from rest at \(x=2\) m, what will be its speed when it reaches \(x=1\) m? (Mass = 2 kg)

1. \(\sqrt{6}\) m/s
2. \(\sqrt{15}\) m/s
3. \(\sqrt{15}\) m/s
4. \(2\sqrt{5}\) m/s

Explanation

\[
F=-\frac{dU}{dx}
\]

Therefore,

\[
U=x^4
\]

Initial energy:

\[
E_i=U(2)=16
\]

At \(x=1\),

\[
U(1)=1
\]

Using conservation of energy,

\[
\frac{1}{2}(2)v^2+1=16
\]

\[
v^2=15
\]

\[
v=\sqrt{15}\;m/s
\]

A satellite is moved from a circular orbit of radius \(R\) to another circular orbit of radius \(4R\). What is the increase in total mechanical energy?

1. \(\dfrac{GMm}{8R}\)
2. \(\dfrac{3GMm}{8R}\)
3. \(\dfrac{3GMm}{8R}\)
4. \(\dfrac{GMm}{4R}\)

Explanation

Initial energy:

\[
E_1=-\frac{GMm}{2R}
\]

Final energy:

\[
E_2=-\frac{GMm}{8R}
\]

Increase:

\[
\Delta E
=
E_2-E_1
\]

\[
=
-\frac{GMm}{8R}
+\frac{GMm}{2R}
\]

\[
=
\frac{3GMm}{8R}
\]

A spring of force constant k is compressed by x and used to launch a block of mass m up a smooth incline of angle \(\theta\). What is the maximum distance traveled along the incline?

1. \(\dfrac{kx^2}{mg\sin\theta}\)
2. \(\dfrac{kx^2}{2mg\sin\theta}\)
3. \(\dfrac{kx^2}{4mg\sin\theta}\)
4. \(\dfrac{2kx^2}{mg\sin\theta}\)

Explanation

Initial spring energy:

\[
\frac{1}{2}kx^2
\]

At the highest point,

\[
mgs\sin\theta
=
\frac{1}{2}kx^2
\]

Thus,

\[
s
=
\frac{kx^2}
{2mg\sin\theta}
\]

A particle moves in one dimension under the potential energy function \(U(x)=x^4-8x^2\). At which position is the particle’s kinetic energy maximum if total energy is conserved?

1. \(x=\pm2\)
2. \(x=\pm2\)
3. \(x=0\)
4. \(x=\pm\sqrt2\)

Explanation

Kinetic energy is maximum where potential energy is minimum.

\[
U=x^4-8x^2
\]

Differentiate:

\[
\frac{dU}{dx}
=
4x(x^2-4)
\]

Critical points:

\[
x=0,\pm2
\]

Checking,

\[
U(\pm2)=-16
\]

which is the minimum value.

Hence kinetic energy is maximum at

\[
x=\pm2
\]

A body of mass m falls freely from a height h. Due to air resistance, 20% of its initial mechanical energy is lost. What is its speed just before reaching the ground?

1. \(\sqrt{gh}\)
2. \(\sqrt{1.2gh}\)
3. \(\sqrt{1.6gh}\)
4. \(\sqrt{2gh}\)

Explanation

Initial energy:

\[
E_i=mgh
\]

20% is lost.

Remaining energy:

\[
0.8mgh
\]

At the ground,

\[
\frac{1}{2}mv^2
=
0.8mgh
\]

\[
v^2
=
1.6gh
\]

\[
v
=
\sqrt{1.6gh}
\]

A force acting on a particle is given by \(F=\dfrac{k}{x^2}\). How much work is done when the particle moves from \(x=a\) to \(x=2a\)?

1. \(\dfrac{k}{a}\)
2. \(\dfrac{k}{4a}\)
3. \(\dfrac{k}{2a}\)
4. \(\dfrac{3k}{2a}\)

Explanation

\[
W
=
\int_a^{2a}
\frac{k}{x^2}dx
\]

\[
=
k
\left(
-\frac{1}{x}
\right)_a^{2a}
\]

\[
=
-\frac{k}{2a}
+\frac{k}{a}
\]

\[
=
\frac{k}{2a}
\]

A particle oscillates in the potential \(U(x)=\dfrac{1}{2}kx^2\). If its total energy is E, what fraction of the total energy is kinetic when \(x=\dfrac{A}{2}\)?

1. \(\dfrac{1}{4}\)
2. \(\dfrac{1}{2}\)
3. \(\dfrac{3}{4}\)
4. \(\dfrac{7}{8}\)

Explanation

Total energy:

\[
E=\frac{1}{2}kA^2
\]

Potential energy at

\[
x=\frac{A}{2}
\]

is

\[
U
=
\frac{1}{2}k
\left(
\frac{A}{2}
\right)^2
=
\frac{E}{4}
\]

Thus,

\[
K
=
E-\frac{E}{4}
=
\frac{3E}{4}
\]

A rocket in deep space ejects fuel. Neglect external forces. Which quantity remains strictly conserved during the process?

1. Kinetic energy
2. Mechanical energy
3. Linear momentum
4. Power

Explanation

In the absence of external forces,

\[
\vec P_{total}
=
constant
\]

Kinetic energy may change because chemical energy is converted into motion.

A particle moves under the force \(F=-kx\). If its amplitude is doubled, how does the total mechanical energy change?

1. Doubles
2. Triples
3. Becomes four times
4. Remains unchanged

Explanation

For SHM,

\[
E
=
\frac{1}{2}kA^2
\]

If

\[
A\rightarrow2A
\]

then

\[
E’
=
\frac{1}{2}k(2A)^2
=
4E
\]

A particle moves in the potential energy field \(U(x)=\dfrac{a}{x^2}\), where \(a>0\). Which statement is correct?

1. The force is attractive
2. The force is constant
3. The force is repulsive and proportional to \(\dfrac{1}{x^3}\)
4. The force is proportional to x

Explanation

\[
F
=
-\frac{dU}{dx}
\]

\[
=
-\left(
-\frac{2a}{x^3}
\right)
\]

\[
=
\frac{2a}{x^3}
\]

The force points away from the origin and varies as

\[
\frac{1}{x^3}
\]

Therefore it is repulsive.