100 Advanced & Numerical MCQs (Level -2) on Work and Energy, Physics (Unit-Wise MCQs Practice):

👉Physics MCQs Hub – Unit wise Physics MCQs Practice (Main-Page)👈

Whether you are preparing for board examinations, chapter tests, college assessments, or competitive entrance exams (MDCAT, ECAT, NUST, PIEAS, GIKI, UET, FAST, and other engineering or medical admission tests), this comprehensive MCQ collection is designed to help you master Work and Energy. The questions are arranged progressively—from basic concepts to advanced numerical problems and higher-order thinking—ensuring complete and systematic preparation for every type of examination.

This chapter-wise MCQ collection includes:

100 Basic MCQs – Covering fundamental concepts, definitions, work, energy, power, SI units, positive, negative and zero work, gravitational field, gravitational potential, conservative and non-conservative forces.

100 Advanced & Numerical MCQs – Focusing on calculations involving work done by constant and variable forces, force-displacement graphs, kinetic and potential energy, work-energy theorem, power, efficiency, gravitational field strength, gravitational potential, and escape velocity.

50 Higher-Order Thinking Skills (HOTS) MCQs – Designed to strengthen analytical reasoning, conceptual understanding, graph interpretation, real-life applications, and multi-concept problem-solving abilities.

This MCQ collection covers:

Work and its calculation using force and displacement
Positive, negative, and zero work
Force-displacement graphs and work calculation
Kinetic energy, gravitational potential energy, and mechanical energy
Gravitational field, field strength, and gravitational potential
Conservative and non-conservative forces
Work done by gravity and conservation of mechanical energy
Escape velocity and its applications
Power, efficiency, and energy transformations
Work-energy theorem in resistive media
Conventional and non-conventional energy sources
Real-life applications and numerical problem-solving

Every MCQ includes the correct answer along with a clear, concept-based explanation to strengthen understanding, improve problem-solving skills, and reinforce important Physics concepts.

This question bank helps students to:

Build a strong conceptual foundation in Work and Energy
Master numerical problem-solving techniques
Understand gravitational field, potential, and escape velocity
Apply the work-energy theorem confidently
Improve analytical reasoning and HOTS skills
Avoid common examination mistakes
Increase speed, accuracy, and confidence in objective-type questions
Prepare effectively for both board examinations and competitive entrance tests

With 250 carefully selected MCQs arranged into 100 Basic, 100 Advanced & Numerical, and 50 HOTS questions, this all-in-one MCQ bank provides complete preparation for Work and Energy. It is an excellent study resource for strengthening concepts, improving exam performance, and achieving success in both school and competitive Physics examinations.

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Level-II – 100 Advanced & Numerical MCQs (MCQs 101–200)

MCQ No. 101

A constant force of 80 N acts on a body and moves it through a displacement of 15 m in the direction of the force. The work done is:

a. 800 J

b. 1000 J

c. 1200 J

d. 1500 J

Correct Answer: c. 1200 J

Explanation:
The work done by a constant force is calculated using

$W = Fd$ $W = 80 \times 15 = 1200 \text{ J}$

MCQ No. 102

A force of 50 N acts at an angle of $60^\circ$ to the direction of displacement. If the body moves 10 m, the work done is:

a. 250 J

b. 500 J

c. 750 J

d. 1000 J

Correct Answer: a. 250 J

Explanation:
When force acts at an angle,

$W = Fd\cos\theta$ $W = 50 \times 10 \times \cos60^\circ$ $W = 500 \times 0.5 = 250\text{ J}$

MCQ No. 103

A force of 120 N pulls a box through 8 m. A frictional force of 20 N opposes the motion. The net work done is:

a. 640 J

b. 800 J

c. 960 J

d. 1120 J

Correct Answer: b. 800 J

Explanation:
Applied work:

$120 \times 8 = 960\text{ J}$

Work done by friction:

$20 \times 8 = 160\text{ J}$

Net work:

$960 - 160 = 800\text{ J}$

MCQ No. 104

A force-displacement graph is a rectangle having a height of 40 N and a width of 12 m. The work done is:

a. 240 J

b. 360 J

c. 480 J

d. 520 J

Correct Answer: c. 480 J

Explanation:
The work done equals the area under the graph.

$W = 40 \times 12 = 480\text{ J}$

MCQ No. 105

A force-displacement graph forms a triangle of base 8 m and height 60 N. The work done is:

a. 120 J

b. 180 J

c. 240 J

d. 300 J

Correct Answer: c. 240 J

Explanation:

$W=\frac12 \times 8 \times 60$ $W=240\text{ J}$

MCQ No. 106

A 5 kg object is lifted vertically through 12 m. Taking $g=10\,\text{m/s}^2$ , the increase in gravitational potential energy is:

a. 300 J

b. 400 J

c. 500 J

d. 600 J

Correct Answer: d. 600 J

Explanation:

$PE=mgh$ $=5\times10\times12 =600\text{ J}$

MCQ No. 107

A body falls freely through a height of 15 m. Ignoring air resistance, the loss in potential energy is:

a. Equal to the gain in kinetic energy

b. Greater than the gain in kinetic energy

c. Less than the gain in kinetic energy

d. Equal to zero

Correct Answer: a. Equal to the gain in kinetic energy

Explanation:
In free fall, mechanical energy is conserved. Therefore, the loss in gravitational potential energy equals the gain in kinetic energy.

MCQ No. 108

A body of mass 4 kg moves with a velocity of 8 m/s. Its kinetic energy is:

a. 64 J

b. 96 J

c. 128 J

d. 256 J

Correct Answer: c. 128 J

Explanation:

$KE=\frac12mv^2$ $=\frac12\times4\times8^2 =2\times64 =128\text{ J}$

MCQ No. 109

If the speed of a moving body is tripled, its kinetic energy becomes:

a. Three times

b. Six times

c. Nine times

d. Twelve times

Correct Answer: c. Nine times

Explanation:
Since

$KE\propto v^2$

tripling the speed increases the kinetic energy by

$3^2=9$

times.

MCQ No. 110

A body initially possesses 250 J of kinetic energy. If the net work done on it is 150 J, its final kinetic energy becomes:

a. 100 J

b. 250 J

c. 400 J

d. 500 J

Correct Answer: c. 400 J

Explanation:
According to the work-energy theorem,

$W=\Delta KE$ $KE_f=250+150=400\text{ J}$

MCQ No. 111

A machine receives 2500 J of energy and delivers 2000 J of useful output. Its efficiency is:

a. 70%

b. 75%

c. 80%

d. 85%

Correct Answer: c. 80%

Explanation:

$\eta=\frac{2000}{2500}\times100 =80\%$

MCQ No. 112

A motor does 6000 J of work in 30 seconds. Its power is:

a. 100 W

b. 150 W

c. 200 W

d. 250 W

Correct Answer: c. 200 W

Explanation:

$P=\frac{W}{t}$ $=\frac{6000}{30} =200\text{ W}$

MCQ No. 113

A force of 90 N acts on an object moving with a velocity of 4 m/s in the same direction. The power developed is:

a. 180 W

b. 270 W

c. 360 W

d. 450 W

Correct Answer: c. 360 W

Explanation:

$P=Fv$ $=90\times4 =360\text{ W}$

MCQ No. 114

A body gains 500 J of gravitational potential energy. How much work is done against gravity?

a. 250 J

b. 500 J

c. 750 J

d. 1000 J

Correct Answer: b. 500 J

Explanation:
The work done against gravity is equal to the increase in gravitational potential energy.

MCQ No. 115

A body is moved from point A to point B through two different paths in Earth's gravitational field. The work done by gravity is:

a. Greater along the longer path

b. Greater along the shorter path

c. Equal for both paths

d. Zero in both cases

Correct Answer: c. Equal for both paths

Explanation:
Gravity is a conservative force. Hence, the work done depends only on the initial and final positions.

MCQ No. 116

The work done by gravity over a complete closed path is:

a. Positive

b. Negative

c. Zero

d. Equal to the body's weight

Correct Answer: c. Zero

Explanation:
For a conservative force like gravity, the total work done over any closed path is always zero.

MCQ No. 117

The escape velocity from a planet is proportional to:

a. $\sqrt{\dfrac{M}{R}}$

b. $\dfrac{M}{R}$

c. $\dfrac{R}{M}$

d. $\sqrt{\dfrac{R}{M}}$

Correct Answer: a. $\sqrt{\dfrac{M}{R}}$

Explanation:
Escape velocity is given by

$v_e=\sqrt{\frac{2GM}{R}}$

Thus,

$v_e\propto\sqrt{\frac{M}{R}}$

MCQ No. 118

If the radius of a planet is increased while its mass remains unchanged, the escape velocity:

a. Increases

b. Decreases

c. Remains constant

d. Becomes zero

Correct Answer: b. Decreases

Explanation:
Since

$v_e\propto\frac1{\sqrt R},$

an increase in radius decreases the escape velocity.

MCQ No. 119

Which of the following represents the work done by a variable force?

a. Product of force and time

b. Area under the force-displacement graph

c. Product of mass and velocity

d. Area under the velocity-time graph

Correct Answer: b. Area under the force-displacement graph

Explanation:
For a variable force, the work done is determined from the area enclosed under the force-displacement graph.

MCQ No. 120

A resistive force removes 180 J of energy from a moving object. The work done by the resistive force is:

a. +180 J

b. -180 J

c. 0 J

d. 1800 J

Correct Answer: b. -180 J

Explanation:
Resistive forces oppose motion and therefore perform negative work, reducing the object's mechanical energy.

MCQ No. 121

A machine has an efficiency of 90%. If the input energy is 4000 J, the useful output energy is:

a. 3200 J

b. 3400 J

c. 3600 J

d. 3800 J

Correct Answer: c. 3600 J

Explanation:

$\eta=\frac{\text{Output}}{\text{Input}}$ $\text{Output}=0.90\times4000=3600\text{ J}$

MCQ No. 122

A force acts perpendicular to the displacement of an object. The work done is:

a. Maximum

b. Positive

c. Negative

d. Zero

Correct Answer: d. Zero

Explanation:

$W=Fd\cos90^\circ=0$

No work is done because the force has no component along the displacement.

MCQ No. 123

A 10 kg object is lifted through a height of 5 m. Taking $g=9.8\,\text{m/s}^2$ , the gain in potential energy is:

a. 245 J

b. 490 J

c. 500 J

d. 980 J

Correct Answer: b. 490 J

Explanation:

$PE=mgh$ $=10\times9.8\times5 =490\text{ J}$

MCQ No. 124

The work-energy theorem is applicable:

a. Only for conservative forces

b. Only in the absence of friction

c. For all forces acting on a body

d. Only for gravitational force

Correct Answer: c. For all forces acting on a body

Explanation:
The work-energy theorem states that the net work done by all forces equals the change in kinetic energy, regardless of whether the forces are conservative or non-conservative.

MCQ No. 125

A body moves under the action of several forces. The algebraic sum of the work done by all these forces is called:

a. Applied work

b. Conservative work

c. Net work

d. Resistive work

Correct Answer: c. Net work

Explanation:
The total or net work is the algebraic sum of the work done by all the forces acting on a body. According to the work-energy theorem, this net work equals the change in the body's kinetic energy.

MCQ No. 126

A force of 150 N acts at an angle of $30^\circ$ to the horizontal and moves an object through 20 m. The work done by the force is approximately:

a. 1500 J

b. 2250 J

c. 2598 J

d. 3000 J

Correct Answer: c. 2598 J

Explanation:
The work done by a force acting at an angle is

$W = Fd\cos\theta$ $W = 150 \times 20 \times \cos30^\circ$ $W = 3000 \times 0.866 \approx 2598\text{ J}$

MCQ No. 127

A 10 kg object slides 25 m on a rough horizontal surface against a frictional force of 18 N. The work done by friction is:

a. +450 J

b. -450 J

c. +180 J

d. -180 J

Correct Answer: b. -450 J

Explanation:
Since friction opposes motion,

$W = -Fd$ $W = -(18)(25) = -450\text{ J}$

MCQ No. 128

A body initially at rest gains 900 J of kinetic energy. According to the work-energy theorem, the net work done on the body is:

a. 0 J

b. 450 J

c. 900 J

d. 1800 J

Correct Answer: c. 900 J

Explanation:
The work-energy theorem states:

$W_{\text{net}}=\Delta KE$

Since the body starts from rest,

$\Delta KE = 900 - 0 = 900\text{ J}$

MCQ No. 129

A body of mass 8 kg is lifted through 15 m. Taking $g=9.8\,\text{m/s}^2$ , the increase in gravitational potential energy is:

a. 980 J

b. 1176 J

c. 1200 J

d. 1470 J

Correct Answer: b. 1176 J

Explanation:

$PE=mgh$ $=8\times9.8\times15 =1176\text{ J}$

MCQ No. 130

A machine has an input power of 1200 W and an output power of 900 W. Its efficiency is:

a. 60%

b. 70%

c. 75%

d. 80%

Correct Answer: c. 75%

Explanation:

$\eta=\frac{900}{1200}\times100 =75\%$

MCQ No. 131

A body moves 50 m under a constant force of 60 N. If the force acts opposite to the displacement, the work done is:

a. +3000 J

b. -3000 J

c. +6000 J

d. 0 J

Correct Answer: b. -3000 J

Explanation:

$W = Fd\cos180^\circ$ $=60\times50\times(-1) =-3000\text{ J}$

MCQ No. 132

The area under a force-displacement graph is 850 J. This value represents:

a. Power

b. Momentum

c. Work done

d. Acceleration

Correct Answer: c. Work done

Explanation:
The area under a force-displacement graph always gives the work done by the force.

MCQ No. 133

A body has a kinetic energy of 450 J. If another 150 J of work is done on it, its final kinetic energy becomes:

a. 300 J

b. 450 J

c. 600 J

d. 750 J

Correct Answer: c. 600 J

Explanation:

$KE_f=450+150 =600\text{ J}$

MCQ No. 134

A force of 100 N moves an object 6 m while making an angle of $60^\circ$ with the displacement. The work done is:

a. 300 J

b. 400 J

c. 500 J

d. 600 J

Correct Answer: a. 300 J

Explanation:

$W=Fd\cos\theta$ $=100\times6\times0.5 =300\text{ J}$

MCQ No. 135

The escape velocity of a planet depends on:

a. The mass of the object

b. The density of the object

c. The planet's mass and radius

d. The shape of the object

Correct Answer: c. The planet's mass and radius

Explanation:
Escape velocity is given by

$v_e=\sqrt{\frac{2GM}{R}}$

It depends only on the planet's mass and radius.

MCQ No. 136

A motor develops 1500 W of power while pulling a load at 5 m/s. The force exerted by the motor is:

a. 150 N

b. 250 N

c. 300 N

d. 500 N

Correct Answer: c. 300 N

Explanation:

$P=Fv$ $F=\frac{P}{v} =\frac{1500}{5} =300\text{ N}$

MCQ No. 137

If the radius of a planet is doubled while its mass remains unchanged, the escape velocity becomes:

a. Double

b. Half

c. $\frac{1}{\sqrt2}$ times

d. Four times

Correct Answer: c. $\frac{1}{\sqrt2}$

Explanation:

Since

$v_e\propto\frac1{\sqrt R},$

doubling the radius gives

$v'_e=\frac{v_e}{\sqrt2}$

MCQ No. 138

A machine receives 5000 J of energy and loses 750 J due to friction. The useful output energy is:

a. 3500 J

b. 4000 J

c. 4250 J

d. 4750 J

Correct Answer: c. 4250 J

Explanation:

Useful output

$=5000-750 =4250\text{ J}$

MCQ No. 139

The work done by gravity when an object is moved between two fixed points depends upon:

a. The path followed

b. The time taken

c. Only the initial and final positions

d. The speed of motion

Correct Answer: c. Only the initial and final positions

Explanation:
Gravity is a conservative force. Therefore, the work done depends only on the starting and ending positions.

MCQ No. 140

A body moves with a constant velocity. The net work done on the body is:

a. Positive

b. Negative

c. Zero

d. Infinite

Correct Answer: c. Zero

Explanation:
When velocity is constant, there is no change in kinetic energy.

According to the work-energy theorem,

$W_{\text{net}}=\Delta KE=0$

MCQ No. 141

A body has a mass of 6 kg and moves at 10 m/s. Its kinetic energy is:

a. 150 J

b. 200 J

c. 300 J

d. 600 J

Correct Answer: c. 300 J

Explanation:

$KE=\frac12mv^2$ $=\frac12\times6\times10^2 =300\text{ J}$

MCQ No. 142

A force-displacement graph is trapezoidal with parallel sides 30 N and 70 N and width 8 m. The work done is:

a. 240 J

b. 320 J

c. 400 J

d. 480 J

Correct Answer: c. 400 J

Explanation:
Area of a trapezium:

$W=\frac12(a+b)h$ $=\frac12(30+70)\times8 =400\text{ J}$

MCQ No. 143

A body is lifted upward slowly. The work done by gravity is:

a. Positive

b. Negative

c. Zero

d. Maximum

Correct Answer: b. Negative

Explanation:
Gravity acts downward while displacement is upward, so the work done by gravity is negative.

MCQ No. 144

A body falls freely from a height of 45 m. Ignoring air resistance, the decrease in potential energy is:

a. Equal to the increase in kinetic energy

b. Less than the increase in kinetic energy

c. Greater than the increase in kinetic energy

d. Zero

Correct Answer: a. Equal to the increase in kinetic energy

Explanation:
Mechanical energy is conserved during free fall in the absence of air resistance.

MCQ No. 145

A machine has an efficiency of 85%. If the useful output is 1700 J, the input energy is:

a. 1800 J

b. 1900 J

c. 2000 J

d. 2200 J

Correct Answer: c. 2000 J

Explanation:

$\eta=\frac{1700}{\text{Input}}$ $\text{Input}=\frac{1700}{0.85} =2000\text{ J}$

MCQ No. 146

The work done by friction on a moving object always results in:

a. Increase in mechanical energy

b. Conversion of mechanical energy into heat

c. Increase in gravitational potential energy

d. Increase in kinetic energy

Correct Answer: b. Conversion of mechanical energy into heat

Explanation:
Friction is a non-conservative force that dissipates mechanical energy as heat.

MCQ No. 147

A body is displaced horizontally while gravity acts vertically downward. The work done by gravity is:

a. Positive

b. Negative

c. Zero

d. Equal to weight

Correct Answer: c. Zero

Explanation:
Gravity is perpendicular to the horizontal displacement.

$W=Fd\cos90^\circ=0$

MCQ No. 148

If the net work done on a body is negative, its kinetic energy:

a. Increases

b. Decreases

c. Remains constant

d. Becomes infinite

Correct Answer: b. Decreases

Explanation:
Negative net work means energy is removed from the body, causing its kinetic energy to decrease.

MCQ No. 149

The work required to move a body from one point to another in a conservative field is independent of:

a. Initial position

b. Final position

c. Path followed

d. Mass of the body

Correct Answer: c. Path followed

Explanation:
In a conservative field, the work done depends only on the initial and final positions, not on the path taken.

MCQ No. 150

A body moves in a gravitational field from point A to point B through three different paths. Which statement is correct?

a. The longest path requires the greatest work by gravity.

b. The shortest path requires the least work by gravity.

c. The work done by gravity is the same for all three paths.

d. The work done depends on the speed of the body.

Correct Answer: c. The work done by gravity is the same for all three paths.

Explanation:
Gravitational force is conservative. Therefore, the work done by gravity depends only on the difference in height (or the initial and final positions) and is completely independent of the path followed.

MCQ No. 151

A body of mass 5 kg is moving with a speed of 12 m/s. If a retarding force does 180 J of work on it, the final kinetic energy of the body is:

a. 90 J

b. 180 J

c. 270 J

d. 360 J

Correct Answer: b. 180 J

Explanation:
Initial kinetic energy:

$KE_i=\frac12mv^2$ $=\frac12\times5\times12^2 =360\text{ J}$

Since the retarding force does −180 J of work,

$KE_f=360-180=180\text{ J}$

MCQ No. 152

A force of 100 N acts on an object for a displacement of 25 m. If the force makes an angle of $37^\circ$ with the displacement ( $\cos37^\circ=0.8$ ), the work done is:

a. 1600 J

b. 1800 J

c. 2000 J

d. 2500 J

Correct Answer: c. 2000 J

Explanation:

$W=Fd\cos\theta$ $=100\times25\times0.8 =2000\text{ J}$

MCQ No. 153

A 20 kg object is raised vertically by 6 m. Taking $g=9.8\,\text{m/s}^2$ , the work done against gravity is:

a. 980 J

b. 1176 J

c. 1200 J

d. 1568 J

Correct Answer: b. 1176 J

Explanation:

$W=mgh$ $=20\times9.8\times6 =1176\text{ J}$

MCQ No. 154

A machine receives 6000 J of energy and has an efficiency of 70%. The useful output energy is:

a. 3600 J

b. 4000 J

c. 4200 J

d. 4800 J

Correct Answer: c. 4200 J

Explanation:

$\eta=\frac{\text{Output}}{\text{Input}}$ $\text{Output}=0.70\times6000 =4200\text{ J}$

MCQ No. 155

A body gains 450 J of kinetic energy while moving over a rough surface where friction does 120 J of negative work. The work done by the applied force is:

a. 330 J

b. 450 J

c. 570 J

d. 690 J

Correct Answer: c. 570 J

Explanation:
Using the work-energy theorem,

$W_{\text{applied}}+W_{\text{friction}}=\Delta KE$ $W_{\text{applied}}-120=450$ $W_{\text{applied}}=570\text{ J}$

MCQ No. 156

The area under a force-displacement graph is 960 J. If the displacement is 12 m, the average force acting on the object is:

a. 60 N

b. 70 N

c. 80 N

d. 90 N

Correct Answer: c. 80 N

Explanation:

$W=Fd$ $F=\frac{960}{12}=80\text{ N}$

MCQ No. 157

A body is projected vertically upward. At the highest point, its:

a. Kinetic energy is maximum.

b. Potential energy is minimum.

c. Kinetic energy is zero.

d. Total mechanical energy is zero.

Correct Answer: c. Kinetic energy is zero.

Explanation:
At the highest point, the velocity becomes zero momentarily. Therefore, kinetic energy becomes zero, while gravitational potential energy is maximum.

MCQ No. 158

A force of 250 N pulls a trolley through 16 m while friction opposes the motion with a force of 50 N. The net work done is:

a. 2400 J

b. 2800 J

c. 3200 J

d. 4000 J

Correct Answer: c. 3200 J

Explanation:

Applied work:

$250\times16=4000\text{ J}$

Work done by friction:

$50\times16=800\text{ J}$

Net work:

$4000-800=3200\text{ J}$

MCQ No. 159

Which statement correctly explains why gravity is called a conservative force?

a. It always acts vertically downward.

b. Its work depends only on the initial and final positions.

c. It always produces positive work.

d. It cannot change kinetic energy.

Correct Answer: b. Its work depends only on the initial and final positions.

Explanation:
The defining property of a conservative force is that the work done depends only on the initial and final positions, not on the path followed.

MCQ No. 160

A body of mass 2 kg has a kinetic energy of 196 J. Its speed is:

a. 10 m/s

b. 12 m/s

c. 14 m/s

d. 16 m/s

Correct Answer: c. 14 m/s

Explanation:

$KE=\frac12mv^2$ $196=\frac12\times2\times v^2$ $v^2=196$ $v=14\text{ m/s}$

MCQ No. 161

If the mass of a planet becomes four times while its radius remains unchanged, its escape velocity becomes:

a. Double

b. Four times

c. Half

d. Unchanged

Correct Answer: a. Double

Explanation:

Since

$v_e=\sqrt{\frac{2GM}{R}}$ $v_e\propto\sqrt{M}$

If the mass becomes four times,

$v_e=\sqrt4=2$

times the original value.

MCQ No. 162

A machine delivers 1500 W of useful power while receiving 2000 W of input power. Its efficiency is:

a. 60%

b. 70%

c. 75%

d. 80%

Correct Answer: c. 75%

Explanation:

$\eta=\frac{1500}{2000}\times100 =75\%$

MCQ No. 163

A constant horizontal force accelerates a body from rest. Which quantity increases because of the work done by the force?

a. Potential energy only

b. Kinetic energy

c. Mass

d. Weight

Correct Answer: b. Kinetic energy

Explanation:
According to the work-energy theorem, the net work done on a body appears as an increase in its kinetic energy.

MCQ No. 164

A body moves in a horizontal circle at constant speed. The work done by the centripetal force is:

a. Positive

b. Negative

c. Zero

d. Maximum

Correct Answer: c. Zero

Explanation:
The centripetal force is always perpendicular to the instantaneous displacement of the body.

Therefore,

$W=Fd\cos90^\circ=0$

MCQ No. 165

A body of mass 10 kg is lifted through 8 m. Taking $g=9.8\,\text{m/s}^2$ , the increase in gravitational potential energy is:

a. 588 J

b. 684 J

c. 784 J

d. 980 J

Correct Answer: c. 784 J

Explanation:

$PE=mgh$ $=10\times9.8\times8 =784\text{ J}$

MCQ No. 166

A body initially has 800 J of kinetic energy. If 300 J of negative work is done on it, its final kinetic energy is:

a. 300 J

b. 400 J

c. 500 J

d. 1100 J

Correct Answer: c. 500 J

Explanation:

$KE_f=800-300 =500\text{ J}$

MCQ No. 167

A body is moved between two points in a gravitational field along three different paths. Which quantity remains the same?

a. Distance travelled

b. Time taken

c. Work done by gravity

d. Speed throughout the motion

Correct Answer: c. Work done by gravity

Explanation:
Since gravity is a conservative force, the work done depends only on the initial and final positions.

MCQ No. 168

A force-displacement graph consists of a rectangle of area 600 J followed by a triangle of area 200 J. The total work done is:

a. 600 J

b. 700 J

c. 800 J

d. 1000 J

Correct Answer: c. 800 J

Explanation:
The total work done equals the total area under the graph.

$600+200=800\text{ J}$

MCQ No. 169

Which statement best describes the work-energy theorem in a resistive medium?

a. Net work equals change in momentum.

b. Net work equals change in potential energy.

c. Net work, including negative work by resistive forces, equals the change in kinetic energy.

d. Friction has no effect on kinetic energy.

Correct Answer: c. Net work, including negative work by resistive forces, equals the change in kinetic energy.

Explanation:
Even in the presence of friction or air resistance, the work-energy theorem remains valid. The net work, including the negative work done by resistive forces, equals the change in kinetic energy.

MCQ No. 170

A machine has an efficiency of 92%. If the input energy is 5000 J, the energy lost is:

a. 200 J

b. 300 J

c. 400 J

d. 500 J

Correct Answer: c. 400 J

Explanation:

Useful output

$=0.92\times5000=4600\text{ J}$

Energy lost

$=5000-4600=400\text{ J}$

MCQ No. 171

The work done by gravity during the upward motion of a projectile is:

a. Positive

b. Negative

c. Zero

d. Infinite

Correct Answer: b. Negative

Explanation:
Gravity acts downward while the projectile moves upward, so the work done by gravity is negative.

MCQ No. 172

A body moving at 15 m/s has a kinetic energy of 450 J. Its mass is:

a. 2 kg

b. 3 kg

c. 4 kg

d. 5 kg

Correct Answer: c. 4 kg

Explanation:

KE=12mv2

$450=\frac12m(15)^2$ $450=112.5m$ $m=4\text{ kg}$

MCQ No. 173

If both the mass and radius of a planet become four times their original values, the escape velocity will:

a. Double

b. Remain unchanged

c. Become half

d. Become four times

Correct Answer: b. Remain unchanged

Explanation:

$v_e\propto\sqrt{\frac{M}{R}}$ $\sqrt{\frac{4M}{4R}}=\sqrt{\frac{M}{R}}$

Hence, the escape velocity remains unchanged.

MCQ No. 174

The negative work done by friction mainly causes:

a. An increase in mechanical energy

b. A decrease in kinetic energy and conversion into heat

c. An increase in gravitational potential energy

d. An increase in escape velocity

Correct Answer: b. A decrease in kinetic energy and conversion into heat

Explanation:
Friction opposes motion and converts mechanical energy into thermal energy, reducing the kinetic energy of the moving object.

MCQ No. 175

A body moves under several forces. The total work done by all the forces is +250 J. According to the work-energy theorem, the kinetic energy of the body:

a. Decreases by 250 J

b. Increases by 250 J

c. Remains unchanged

d. Becomes zero

Correct Answer: b. Increases by 250 J

Explanation:
According to the work-energy theorem,

$W_{\text{net}}=\Delta KE$

A positive net work of 250 J means the body's kinetic energy increases by 250 J.

MCQ No. 176

A force-displacement graph is a rectangle of height 100 N and width 15 m. The work done by the force is:

a. 1000 J

b. 1200 J

c. 1500 J

d. 1800 J

Correct Answer: c. 1500 J

Explanation:
The work done is equal to the area under the force-displacement graph.

$W = F \times d$ $W = 100 \times 15 = 1500\text{ J}$

MCQ No. 177

A 12 kg object is lifted through 25 m. Taking $g = 9.8\,\text{m/s}^2$ , the increase in gravitational potential energy is:

a. 2450 J

b. 2646 J

c. 2940 J

d. 3000 J

Correct Answer: c. 2940 J

Explanation:

$PE = mgh$ $=12 \times 9.8 \times 25 =2940\text{ J}$

MCQ No. 178

A machine has an efficiency of 88%. If it receives 5000 J of input energy, the useful output energy is:

a. 4200 J

b. 4300 J

c. 4400 J

d. 4500 J

Correct Answer: c. 4400 J

Explanation:

$\eta=\frac{\text{Output}}{\text{Input}}\times100$ $\text{Output}=0.88\times5000=4400\text{ J}$

MCQ No. 179

A body moving with a kinetic energy of 900 J experiences a resistive force that performs 250 J of negative work. Its final kinetic energy is:

a. 550 J

b. 650 J

c. 750 J

d. 1150 J

Correct Answer: b. 650 J

Explanation:

According to the work-energy theorem,

$KE_f=900-250=650\text{ J}$

MCQ No. 180

A body of mass 4 kg moves with a speed of 15 m/s. Its kinetic energy is:

a. 300 J

b. 400 J

c. 450 J

d. 500 J

Correct Answer: c. 450 J

Explanation:

$KE=\frac12mv^2$ $=\frac12\times4\times15^2 =2\times225 =450\text{ J}$

MCQ No. 181

A body moves under the action of gravity from height A to height B through two different paths. The work done by gravity is:

a. Greater along the shorter path

b. Greater along the longer path

c. Equal for both paths

d. Zero for both paths

Correct Answer: c. Equal for both paths

Explanation:
Gravitational force is conservative. Therefore, the work done depends only on the initial and final heights and not on the path followed.

MCQ No. 182

A motor develops a power of 2400 W while pulling an object at 8 m/s. The pulling force is:

a. 200 N

b. 250 N

c. 300 N

d. 350 N

Correct Answer: c. 300 N

Explanation:

$P=Fv$ $F=\frac{P}{v} =\frac{2400}{8} =300\text{ N}$

MCQ No. 183

If the radius of a planet is reduced to one-fourth while its mass remains unchanged, the escape velocity becomes:

a. Half

b. Double

c. Four times

d. Unchanged

Correct Answer: b. Double

Explanation:

$v_e\propto\frac1{\sqrt R}$

$R'=\frac R4$

then

$v'_e=\sqrt4\,v_e=2v_e$

MCQ No. 184

A body gains 980 J of gravitational potential energy. The work done against gravity is:

a. 490 J

b. 980 J

c. 1470 J

d. 1960 J

Correct Answer: b. 980 J

Explanation:
The work done against gravity is equal to the increase in gravitational potential energy.

MCQ No. 185

A force of 180 N acts at an angle of $60^\circ$ and moves an object through 20 m. The work done is:

a. 1800 J

b. 2400 J

c. 3000 J

d. 3600 J

Correct Answer: a. 1800 J

Explanation:

$W=Fd\cos\theta$ $=180\times20\times0.5 =1800\text{ J}$

MCQ No. 186

A machine loses 12% of its input energy due to friction. Its efficiency is:

a. 82%

b. 85%

c. 88%

d. 90%

Correct Answer: c. 88%

Explanation:
Efficiency

$=100\%-12\% =88\%$

MCQ No. 187

A body moves with constant speed on a rough horizontal surface. The applied force is mainly used to:

a. Increase kinetic energy

b. Increase potential energy

c. Balance frictional losses

d. Increase momentum continuously

Correct Answer: c. Balance frictional losses

Explanation:
Since the speed is constant, the kinetic energy does not change. The applied force only compensates for the work done against friction.

MCQ No. 188

The work done by a conservative force over any closed path is always:

a. Positive

b. Negative

c. Zero

d. Maximum

Correct Answer: c. Zero

Explanation:
For conservative forces such as gravity, the total work done over any closed path is always zero.

MCQ No. 189

A body of mass 10 kg is moving at 20 m/s. Its kinetic energy is:

a. 1000 J

b. 1500 J

c. 2000 J

d. 2500 J

Correct Answer: c. 2000 J

Explanation:

$KE=\frac12mv^2$ $=\frac12\times10\times20^2 =2000\text{ J}$

MCQ No. 190

A force-displacement graph is triangular with base 12 m and height 150 N. The work done is:

a. 600 J

b. 750 J

c. 900 J

d. 1200 J

Correct Answer: c. 900 J

Explanation:

$W=\frac12\times12\times150 =900\text{ J}$

MCQ No. 191

A body is projected vertically upward with an initial kinetic energy of 800 J. Ignoring air resistance, its gravitational potential energy at the highest point will be:

a. 200 J

b. 400 J

c. 600 J

d. 800 J

Correct Answer: d. 800 J

Explanation:
In the absence of air resistance, mechanical energy is conserved. At the highest point, all the initial kinetic energy is converted into gravitational potential energy.

MCQ No. 192

The net work done on a body is -500 J. This indicates that the body's kinetic energy:

a. Increases by 500 J

b. Decreases by 500 J

c. Remains constant

d. Doubles

Correct Answer: b. Decreases by 500 J

Explanation:
According to the work-energy theorem,

$W_{\text{net}}=\Delta KE$

A negative value means the kinetic energy decreases.

MCQ No. 193

A force of 75 N acts on a body moving at 12 m/s in the same direction. The instantaneous power developed is:

a. 750 W

b. 800 W

c. 900 W

d. 1000 W

Correct Answer: c. 900 W

Explanation:

$P=Fv$ $=75\times12 =900\text{ W}$

MCQ No. 194

The work required to move a body from one point to another in a gravitational field depends upon:

a. Time taken

b. Distance travelled

c. Difference in position only

d. Speed of the body

Correct Answer: c. Difference in position only

Explanation:
Since gravity is a conservative force, the work done depends only on the initial and final positions.

MCQ No. 195

A body receives 1200 J of net work. If its initial kinetic energy is 500 J, its final kinetic energy is:

a. 700 J

b. 1200 J

c. 1700 J

d. 2200 J

Correct Answer: c. 1700 J

Explanation:

$KE_f=500+1200 =1700\text{ J}$

MCQ No. 196

The gravitational field strength at a point is numerically equal to:

a. Potential energy per unit mass

b. Force acting on unit mass

c. Work done per unit distance

d. Energy per unit volume

Correct Answer: b. Force acting on unit mass

Explanation:
Gravitational field strength is defined as the gravitational force experienced by a unit mass placed at a given point.

MCQ No. 197

A machine delivers 5400 J of useful energy with an efficiency of 90%. The input energy supplied to the machine is:

a. 5800 J

b. 5900 J

c. 6000 J

d. 6200 J

Correct Answer: c. 6000 J

Explanation:

$\eta=\frac{5400}{\text{Input}}$ $\text{Input}=\frac{5400}{0.90} =6000\text{ J}$

MCQ No. 198

Which statement correctly explains why friction is called a non-conservative force?

a. It always acts downward.

b. Its work depends upon the path followed.

c. It acts only on moving bodies.

d. It produces only positive work.

Correct Answer: b. Its work depends upon the path followed.

Explanation:
The work done by friction depends on the distance travelled. Therefore, unlike gravity, friction is a non-conservative force.

MCQ No. 199

The work done by a variable force is determined by calculating the:

a. Slope of the force-displacement graph

b. Area under the force-displacement graph

c. Slope of the velocity-time graph

d. Area under the acceleration-time graph

Correct Answer: b. Area under the force-displacement graph

Explanation:
For a variable force, the work done is equal to the area enclosed between the force-displacement curve and the displacement axis.

MCQ No. 200

A body starts from rest and gains 1600 J of kinetic energy while moving on a rough surface. During the motion, friction does 400 J of negative work. The work done by the applied force is:

a. 1200 J

b. 1600 J

c. 1800 J

d. 2000 J

Correct Answer: d. 2000 J

Explanation:
According to the work-energy theorem,

$W_{\text{applied}}+W_{\text{friction}}=\Delta KE$ $W_{\text{applied}}-400=1600$ $W_{\text{applied}}=2000\text{ J}$

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Work and Energy – 100 Advanced & Numerical MCQs with Solutions | ECAT, MDCAT & Engineering Entry Tests

100 Advanced & Numerical MCQs (Level -2) on Work and Energy, Physics (Unit-Wise MCQs Practice):

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Level-II – 100 Advanced & Numerical MCQs (MCQs 101–200)

MCQ No. 101

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MCQ No. 168

MCQ No. 169

MCQ No. 170

MCQ No. 171

MCQ No. 172

MCQ No. 173

MCQ No. 174

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