Work Power and Energy MCQ Quiz - Objective Question with Answer for Work Power and Energy - Download Free PDF
Last updated on May 12, 2025
Latest Work Power and Energy MCQ Objective Questions
Work Power and Energy Question 1:
A body of 20 kg is lying at rest. Under the action of a constant force, it gains a speed of 7 m/s. The work done by the force will be _______.
Answer (Detailed Solution Below)
Work Power and Energy Question 1 Detailed Solution
The correct answer is 490J
CONCEPT:
- Work-energy theorem: It states that the sum of work done by all the forces acting on a body is equal to the change in the kinetic energy of the body i.e.,
Work done by all the forces = Kf - Ki
\(W = \frac{1}{2}m{v^2} - \frac{1}{2}m{u^2} = {\bf{Δ }}K\)
Where v = final velocity, u = initial velocity and m = mass of the body
CALCULATION:
It is given that,
Mass (m) = 20 kg
Final Velocity (v) = 7 m/s and initial velocity (u) = 0 m/s
According to the work-energy theorem,
⇒ Work done = Change in K.E
⇒ W = Δ K.E
Since initial speed is zero so the initial KE will also be zero.
⇒ Work done (W) = Final K.E = 1/2 mv2
⇒ W = 1/2 × 20 × 72
⇒ W = 10 × 49
⇒ W = 490J
Work Power and Energy Question 2:
What will be the energy possessed by a stationary object of mass 10 kg placed at a height of 20 m above the ground? (take g = 10 m/s2)
Answer (Detailed Solution Below)
Work Power and Energy Question 2 Detailed Solution
The correct answer is 2 kJ.
CONCEPT:
- Potential energy: The energy of any object due to its position with respect to a reference point is called potential energy. It is denoted by PE.
Potential energy is given by:
PE = m g h.
Here, PE is the Potential Energy, m is the mass, g is the acceleration due to gravity, and h is the height at which the object is placed
CALCULATION:
Given that:
Mass (m) = 10 Kg
Height (h) = 20 m
P.E. = 10 x 10 x 20
P.E.= 2000 J
P.E. = 2 kJ
- Kinetic energy: The energy due to the motion of the object is called kinetic energy.
- Kinetic energy (KE) = 1/2 (mv2)
- Where m is mass and v is velocity.
- Since the object is stationary (at rest) so the velocity is zero. Hence the kinetic energy of the object will be zero.
- Only the potential energy of the object will be there at the height.
Work Power and Energy Question 3:
A particle of unit mass is moving along the x-axis under the influence of a force and its total energy is conserved. Four possible forms of the potential energy of the particle are given in Column I (a and U₀ are constants). Match the potential energies in Column I to the corresponding statement(s) in Column II.
Column I | Column II |
---|---|
(A) U₁(x) = (U₀ / 2) [1 - (x² / a²)]² | (p) Force on the particle is zero at x = a |
(B) U₂(x) = (U₀ / 2) (x² / a²) | (q) Force on the particle is zero at x = 0 |
(C) U₃(x) = (U₀ / 2) (x² / a²) exp[ - (x² / a²)] | (r) Force on the particle is zero at x = -a |
(D) U₄(x) = (U₀ / 2) [ (x / a) - (1/3) (x / a)³ ] | (s) The particle experiences attractive force towards x = 0 in |x| < a |
(t) The particle with total energy U₀ / 4 can oscillate about x = -a |
Answer (Detailed Solution Below)
A → (p, q, r, t), B → (q, s), C → (p, q, r, s), D → (p, r, t)
Work Power and Energy Question 3 Detailed Solution
Calculation:
(A) Force is zero at x = -a, 0, a. Oscillation about x = ±a for energy U₀ / 4 → (p), (q), (r), (t)
(B) Force is zero at x = 0. It's attractive near x = 0 → (q), (s)
(C) Force is zero at x = -a, 0, a. Potential well around x = 0 → (p), (q), (r), (s)
(D) Force is zero at x = -a and x = a. Oscillation possible around x = -a for U₀ / 4 → (p), (r), (t)
Work Power and Energy Question 4:
Two identical balls A and B, each of mass 0.1 kg, are attached to two identical massless springs. The spring-mass system is constrained to move inside a rigid smooth pipe bent in the form of a circle as shown in the figure. The pipe is fixed in a horizontal plane. The centres of the balls can move in a circle of radius 0.06 m. Each spring has a natural length of 0.06π m and spring constant 0.1 N/m. Initially, both the balls are displaced by an angle θ = π/6 rad with respect to the diameter PQ of the circle and released from rest. The total energy of the system is α × 10-n J. The ratio of n/α is closet to (nearest integer)
Answer (Detailed Solution Below) 1
Work Power and Energy Question 4 Detailed Solution
Calculation:
The length l is equal to half of the circle's perimeter: l = πR. Thus, the equilibrium positions of the balls A and B are at P and Q, respectively.
Let θ be the angular displacement of each ball from its equilibrium position. In this condition, one spring is compressed by length Rθ and the other is extended by the same length. So the restoring force on ball B by the two springs is:
F = k(2Rθ) + k(2Rθ) = 4kRθ
The tangential acceleration at of the ball B is:
at = R × d²θ/dt²
Apply Newton's second law:
mR × d²θ/dt² = -4kRθ
⇒ d²θ/dt² = -(4k/m)θ
This is the equation of SHM with angular frequency ω² = 4k/m
Frequency, ν = (1 / 2π) × √(4k / m) = (1 / 2π) × √(0.4 / 0.1) = 1 / π Hz
At the initial position (θ = π/6), both springs are stretched/compressed by Rθ. Potential energy of each spring:
U1,i = U2,i = (1/2) × k × (2Rθ)² = 2kR²θ²
Total initial potential energy of system:
Ui = 4kR²θ²
Initial kinetic energy = 0
Final position: each spring is relaxed, so Uf = 0
Total kinetic energy at final position:
Kf = m × v² (since both balls move at same speed)
Using conservation of energy:
Ui + Ki = Uf + Kf
4kR²θ² = m × v²
Solving for v:
v = 2Rθ × √(k / m) = 2 × 0.06 × (π / 6) × √(0.1 / 0.1) = 0.0628 m/s
Total energy of the system:
E = Ui = Kf = 4kR²θ²
E = 4 × 0.1 × (0.06)² × (π / 6)² = 3.9 × 10⁻⁴ J
Thus n / α = 4/ 3.9 ≈ 1
Work Power and Energy Question 5:
A 20 g bullet pierces through a plate of mass m1 = 1 kg and then comes to rest inside a second plate of mass m2 = 2.98 kg as shown in the figure. It is found that the two plates initially at rest now move with equal velocities. Find the percentage loss in the initial velocity of the bullet when it is between m1 and m2. Neglect any loss of material of the plates due to the action of the bullet. Both plates are lying on a smooth table.
Answer (Detailed Solution Below) 25
Work Power and Energy Question 5 Detailed Solution
Calculation:
The bullet of mass m = 20 g = 0.02 kg first collides with a plate of mass m1 = 1 kg and then with a second plate of mass m2 = 2.98 kg.
Let u be the initial velocity of the bullet, and v be its velocity after passing through m1. Let v1 be the velocity of m1 after the bullet passes through it.
Apply conservation of linear momentum before and after first collision:
mu = mv + m1v1 (1)
Now, the bullet embeds into the second plate m2, so they move with common velocity v1 (since all move together after collision).
Apply conservation of linear momentum again before and after second collision:
mv + m1v1 = (m + m2)v1 (2)
Now eliminate v1 from equations (1) and (2):
v / u = (m + m2) / (m + m1 + m2)
= (0.02 + 2.98) / (0.02 + 1 + 2.98) = 3.00 / 4.00 = 0.75
The bullet retains 75% of its velocity. So, percentage loss in velocity = 25%
Answer: 25%
Top Work Power and Energy MCQ Objective Questions
What will be the energy possessed by a stationary object of mass 10 kg placed at a height of 20 m above the ground? (take g = 10 m/s2)
Answer (Detailed Solution Below)
Work Power and Energy Question 6 Detailed Solution
Download Solution PDFThe correct answer is 2 kJ.
CONCEPT:
- Potential energy: The energy of any object due to its position with respect to a reference point is called potential energy. It is denoted by PE.
Potential energy is given by:
PE = m g h.
Here, PE is the Potential Energy, m is the mass, g is the acceleration due to gravity, and h is the height at which the object is placed
CALCULATION:
Given that:
Mass (m) = 10 Kg
Height (h) = 20 m
P.E. = 10 x 10 x 20
P.E.= 2000 J
P.E. = 2 kJ
- Kinetic energy: The energy due to the motion of the object is called kinetic energy.
- Kinetic energy (KE) = 1/2 (mv2)
- Where m is mass and v is velocity.
- Since the object is stationary (at rest) so the velocity is zero. Hence the kinetic energy of the object will be zero.
- Only the potential energy of the object will be there at the height.
A body of 20 kg is lying at rest. Under the action of a constant force, it gains a speed of 7 m/s. The work done by the force will be _______.
Answer (Detailed Solution Below)
Work Power and Energy Question 7 Detailed Solution
Download Solution PDFThe correct answer is 490J
CONCEPT:
- Work-energy theorem: It states that the sum of work done by all the forces acting on a body is equal to the change in the kinetic energy of the body i.e.,
Work done by all the forces = Kf - Ki
\(W = \frac{1}{2}m{v^2} - \frac{1}{2}m{u^2} = {\bf{Δ }}K\)
Where v = final velocity, u = initial velocity and m = mass of the body
CALCULATION:
It is given that,
Mass (m) = 20 kg
Final Velocity (v) = 7 m/s and initial velocity (u) = 0 m/s
According to the work-energy theorem,
⇒ Work done = Change in K.E
⇒ W = Δ K.E
Since initial speed is zero so the initial KE will also be zero.
⇒ Work done (W) = Final K.E = 1/2 mv2
⇒ W = 1/2 × 20 × 72
⇒ W = 10 × 49
⇒ W = 490J
An object with a mass of 22 kg moving with a velocity of 5 m/s possesses kinetic energy of:
Answer (Detailed Solution Below)
Work Power and Energy Question 8 Detailed Solution
Download Solution PDFCONCEPT:
- Kinetic energy: The energy needed to move the body of mass m from one point to another with stated velocity v is called kinetic energy.
The Kinetic energy is given as:
K.E = ½ × m V2
Where K.E = Kinetic Energy
m = mass of the object
V = Velocity of an object
CALCULATION:
Given that, m = 22 kg, v = 5 m/s
∴ K.E = ½ × 22 × 52
K.E = 275 J
Therefore, the kinetic energy of the object is 275 JA man increases the speed of his car from 10 m/s to 20 m/s on level road. The ratio of the final kinetic energy to initial kinetic energy is:
Answer (Detailed Solution Below)
Work Power and Energy Question 9 Detailed Solution
Download Solution PDFCONCEPT:
- Kinetic energy (KE): The energy possessed by a body by virtue of its motion is called kinetic energy.
\(KE = \frac{1}{2}m{v^2}\)
Where m = mass of the body and v = velocity of the body
CALCULATION:
Given that:
Initial speed (u) = 10 m/s
Final speed (v) = 20 m/s
Let the total mass is m.
\(KE = \frac{1}{2}m{v^2}\)
The ratio of final KE to initial KE is given by:
\(Ratio = \frac{\frac{1}{2}mv^2}{\frac{1}{2}mu^2} = \frac{v^2}{u^2} =\frac{20^2}{10^2}=4:1\)
The work done by a person in carrying a box of mass 15 kg through a vertical height of 5 m is 5000 J. The mass of the person is _______ Take g = 10 m/s2
Answer (Detailed Solution Below)
Work Power and Energy Question 10 Detailed Solution
Download Solution PDFConcept:
- Potential energy is defined as the energy stored due to change in position relative to others, stresses within itself, or many factors.
- The potential energy (U) = m g h [where m= mass of body, g= acceleration due to gravity, h = distance from the ground].
- If the height of a body increases from the ground its energy also increases and vice versa.
Calculation:
Work-done to carry a box up to height = m g h
Here the person himself do work on his own mass + mass of box
Let the mass of the person be x
So,
Work-done by potential energy = mass (both) × g × height
⇒ 5000 J = (x kg + 15 kg) × 10 × 5
⇒ 5000 = (x+15) × 50
⇒ 100 = x + 15
⇒ x = 85 kg
Hence the mass of the person is 85 kg.
A train moving with a velocity of 30 km/h has a kinetic energy of 52000 J. When the velocity of train is increased to 60 km/h, the work done is:
Answer (Detailed Solution Below)
Work Power and Energy Question 11 Detailed Solution
Download Solution PDFCONCEPT:
- Work-energy theorem: It states that the sum of work done by all the forces acting on a body is equal to the change in the kinetic energy of the body i.e.,
Work done by all the forces = Kf - Ki
\(W = \frac{1}{2}m{v^2} - \frac{1}{2}m{u^2} = {\bf{Δ }}K\)
Where v = final velocity, u = initial velocity and m = mass of the body
CALCULATION:
It is given that,
Initial velocity (u) = 30 km/h = (30 × 1000/3600) = 25/3 m/s
Initial Kinetic energy (KEi) = 52000 = \(\frac{1}{2}mu^2\)
Final Velocity (v) = 60 km/h = (60 × 1000/3600) = 50/3 m/s = 2u
Final kinetic energy (KEf) = \(\frac{1}{2}mv^2 = \frac{1}{2}m (2u)^2 =4 KE_i \)
⇒ KEf = 4 × 52000 = 208000 J
- According to the work-energy theorem,
⇒ Work done = Change in K.E
⇒ Work done (W) = Δ K.E = KEf - KEi = 208000 - 52000 = 156000 J
A car of mass 1000 kg is moving with velocity 15 ms-1 on a horizontal plane. Work done by the force of gravity will be (g = 10 ms-2)
Answer (Detailed Solution Below)
Work Power and Energy Question 12 Detailed Solution
Download Solution PDFCONCEPT:
- Work is said to be done by a force when the body is displaced actually through some distance in the direction of the applied force.
- Since the body is being displaced in the direction of F, therefore work done by the force in displacing the body through a distance s is given by:
\(W = \vec F \cdot \vec s\)
Or, W = Fs cos θ
- Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body.
- Work done is positive when the direction of both force and displacement are the same
- Work done is negative when the direction of force and displacement are the opposite.
- Whereas work is done is zero when direction force (Centripetal force) and displacement are perpendiculars this case is true for an object moving in a circular motion
CALCULATION:
Given that:
Mass of the car (m) = 1000 kg
Velocity (v) = 15 m/s
- The car is moving on the horizontal plane, so the displacement is in a horizontal direction.
- The gravity force (equal to the weight of the car) is acting in a vertically downward direction.
- So the angle between the force (weight) and the displacement is 90°.
Work done (W) = Fs Cos90° = 0
So option 1 is correct.
A force of 12 N displaces a body by 60 cm in its direction. The work done on the body will be
Answer (Detailed Solution Below)
Work Power and Energy Question 13 Detailed Solution
Download Solution PDFCONCEPT:
- Work is said to be done by a force when the body is displaced actually through some distance in the direction of the applied force.
- Since the body is being displaced in the direction of F, therefore work done by the force in displacing the body through a distance s is given by:
\(W = \vec F \cdot \vec s\)
Or, W = Fs cos θ
- Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body.
CALCULATION:
Given that:
Force (F) = 12 N
Displacement (s) = 60 cm = 60/100 = 0.6 m
The force (F) and displacement (s) are in the same direction, so θ = 0°
Work done = Fs cos θ = 12 × 0.6 × 1 = 7.2 J
Hence option 2 is correct.
An object of mass 20 kg is lifted to a height of 5 m in 20 sec. The power required will be: (g=10m/s2)
Answer (Detailed Solution Below)
Work Power and Energy Question 14 Detailed Solution
Download Solution PDFCONCEPT:
Power:
- The rate of doing work is called power.
- SI unit of power is the watt.
\(⇒ P=\frac{W}{t}\)
.Where P = power, W = work done and t = time
CALCULATION:
Given m = 20 kg, h = 5 m, t = 20 sec and g = 10 m/sec2
- We know that the work done in lifting a body of mass m to height h is given as,
⇒ W = mgh -----(1)
By equation 1,
⇒ W = 20 × 10 × 5 = 1000 J
- So power required,
\(⇒ P=\frac{W}{t}\)
\(⇒ P=\frac{1000}{20}\)
\(⇒ P=50\,watt\)
- Hence, option 1 is correct.
When a force of 50 N is applied on an object of 5 kg, it covers a horizontal distance of 5 m. Calculate the amount of work done in this case.
Answer (Detailed Solution Below)
Work Power and Energy Question 15 Detailed Solution
Download Solution PDFCONCEPT:
- Work is said to be done by a force when the body is displaced actually through some distance in the direction of the applied force.
- Since the body is being displaced in the direction of F, therefore work done by the force in displacing the body through a distance s is given by
\(W = \vec F \cdot \vec s\)
Or, W = Fs cos θ
- Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body.
CALCULATION:
Given that:
Force (F) = 50 , Displacement (s) = 5 m and θ = 0 (because the force is in the direction of motion)
To find the amount of work done in this case, we must apply the following formula:
⇒ W = Fs cosθ
⇒ W = 50 × 5 = 250 J