Work Done by Constant and Variable Forces - NEET Physics Questions
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Work Done by Constant and Variable Forces

Question 21: easy

Assertion (A): There is no term like instantaneous work similar to instantaneous velocity.


Reason (R): For work to be done, the force must act for a displacement.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is true: Velocity is a rate of change at an instant, but work \(W = \int F \cdot dr\) fundamentally involves displacement.


Instantaneous power \(P = F \cdot v\) exists, but not instantaneous work.


Reason (R) is true: Work requires a force to act over a non-zero displacement. Reason (R) correctly explains why instantaneous work is not a valid concept.

Question 22: easy

Assertion (A): A man of mass \(m\), standing on a frictionless surface pushes a wall and acquires a velocity \(v_0\). The work done by the wall on the man is non-zero.


Reason (R): Work done by all the forces is equal to change in kinetic energy.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is true: The wall exerts a reaction force on the man, which accelerates him to velocity \(v_0\). Since the man undergoes displacement while this force acts, the work done by the wall on the man is positive and non-zero, increasing his kinetic energy. Reason (R) is true: This is the Work-Energy Theorem (\(W_{net} = \Delta KE\)). Reason (R) correctly explains why the work done is non-zero as the man gains kinetic energy.

Question 23: easy

Assertion (A): Power delivered by all forces acting on a particle moving in a uniform circular motion is always zero.


Reason (R): Work done by all forces acting on a particle moving in a uniform circular motion is zero as KE remains constant.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is true: In uniform circular motion, the net force (centripetal force) is always perpendicular to the velocity. Power \(P = F \cdot v = |F||v| \cos 90^\circ = 0\). Reason (R) is true: Since speed is constant, kinetic energy \(KE\) is constant, thus \(Delta KE = 0\). By the Work-Energy Theorem, net work done \(W_{net} = \Delta KE = 0\). Reason (R) correctly explains why power is zero.

Question 24: easy

Assertion (A): A man carrying a load on his head and walking with uniform velocity on a street does not work against gravity.


Reason (R): When a body moves with uniform velocity, work done by all forces on this body is zero.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Assertion (A) is true because the displacement (horizontal) is perpendicular to the gravitational force (vertical), so work done \(W = \vec{F} \cdot \vec{d} = Fd \cos(90^{circ}) = 0\).


Reason (R) is also true by the Work-Energy Theorem for uniform velocity. However, (R) does not explain (A).

Question 25: easy

Assertion (A): The kinetic energy of a particle continuously increases with time if the resultant force on the particle must be at an angle less than \(90^{\circ}\) to the velocity at all instants.


Reason (R): The work done by the external forces on a system equals to change in total energy.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Assertion (A) is true.


Kinetic energy \(K\) increases if the net work done \(W_{\text{net}}\) is positive. Since \(W_{\text{net}} = \int \vec{F} \cdot d\vec{s}\) and \(d\vec{s}\) is in the direction of \(vec{v}\) , \(W_{\text{net}} > 0\) implies \(vec{F} \cdot \vec{v} > 0\). This means the angle between \(vec{F}\) and \(vec{v}\) must be less than \(90^{circ}\).


Reason (R) is also true, interpreting "total energy" as kinetic energy in the context of \(W_{\text{net}} = \Delta K\) for a particle or total mechanical energy for a system with external work. Reason (R) provides the fundamental principle behind Assertion (A). Therefore, both are true and R explains A.

Question 26: easy

When a spring is stretched by 1 cm, it stores energy 50 J. If it is further stretched by 1 cm, the stored energy will be

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

Energy is \(U = \frac{1}{2} k x^2\). Since \(U \propto x^2\), doubling the stretch from \(1\text{ cm}\) to \(2\text{ cm}\) increases the stored energy by a factor of \(2^2 = 4\). Thus, \(U' = 4 \times 50\text{ J} = 200\text{ J}\).

Question 27: difficult

A position dependent force, \(F = (7 – 2x + 3x^2)\text{ N})\) acts on a small body of mass \(2\text{ kg})\) and displaces it from \(x = 0\) to \(x = 5\text{ m})\). The work done in joule is:

(1994, 92)

1. 135
2. 270
3. 35
4. 70
View Answer

\(W = int F dx\). Given \(F = 7 - 2x + 3x^2\). Limits \(x = 0\) to \(x = 5\text{ m})\). \(W = [7x - x^2 + x^3]_0^5 = 7(5) - 5^2 + 5^3 = 35 - 25 + 125 = 135\text{ J}\).

Question 28: moderate

A force \(F = 20 + 10y\) acts on a particle in \(y\) direction where \(F\) is in newton and \(y\) in meter. Work done by this force to move the particle from \(y = 0\) to \(y = 1\text{ m})\) is:

(2019)

1. 30 J
2. 5 J
3. 25 J
4. 20 J
View Answer

Work \(W = \int F dy\). Given \(F = 20 + 10y\). Limits \(y = 0\) to \(y = 1\text{ m})\). \(W = \int_0^1 (20 + 10y) dy = [20y + 5y^2]_0^1 = 20(1) + 5(1)^2 = 25\text{ J}\).

Question 29: moderate

A particle moves from a point \(P_1 = (-2\hat{i} + 5\hat{j})\) to \(P_2 = (4\hat{j} + 3\hat{k})\) when a force of \(F = (4\hat{i} + 3\hat{j})\text{ N})\) is applied. How much work has been done by the force?

(2016 – II)

1. 5 J
2. 2 J
3. 8 J
4. 11 J
View Answer

Displacement \(dr = \vec{r_2} - \vec{r_1} = (2\hat{i} - \hat{j} + 3\hat{k})\). Force \(F = (4\hat{i} + 3\hat{j})\). Work \(W = F \cdot dr\). \(W = (4)(2) + (3)(-1) = 8 - 3 = 5\text{ J}\).

Question 30: moderate

A uniform force of \(F = (3\hat{i} + \hat{j})\text{ N})\) acts on a particle of mass \(2\text{ kg})\). Hence the particle is displaced from position \(\vec{r_1} = (2\hat{i} + \hat{k})\text{ m})\) to position \(\vec{r_2} = (4\hat{i} + 3\hat{j} – \hat{k})\text{ m})\). The work done by the force on the particle is:

(2013)

1. 15 J
2. 9 J
3. 6 J
4. 13 J
View Answer

Displacement \(dr = \vec{r_2} - \vec{r_1} = (2\hat{i} + 3\hat{j} - 2\hat{k})\). Force \(F = (3\hat{i} + \hat{j})\). Work \(W = F \cdot dr\). \(W = (3)(2) + (1)(3) = 6 + 3 = 9\text{ J}\).