Work Power and Energy
125 Questions
Start Objective Test
Q101
Objective
1. Taking it together
MCQ
A particle is released from a height $H$ . At certain height, its kinetic energy is two times its potential energy. Height and speed of particle at that instant are
A.
$\frac{H}{3}, \sqrt{\frac{2gH}{3}}$
B.
$\frac{H}{3}, 2\sqrt{\frac{gH}{3}}$
C.
$\frac{2H}{3}, \sqrt{\frac{2gH}{3}}$
D.
$\frac{H}{3}, \sqrt{2gH}$
Q102
Objective
1. Taking it together
MCQ
A force of $F = 0.5 \, \text{N}$ is applied on lower block as shown in figure . The work done by lower block on upper block for a displacement of $3 \, \text{m}$ of the upper block with respect to ground is (Take, $g = 10 \, \text{ms}^{-2}$ )
A.
$-0.5\mathrm{J}$
B.
$0.5\mathrm{J}$
C.
$2\mathrm{J}$
D.
$-2\mathrm{J}$
Q103
Objective
1. Taking it together
MCQ
A body is moving unidirectionally under the influence of a source of constant power supplying energy. Which of the following diagrams shown in figure correctly represents the displacement-time curve for its motion? [NCERT Exemplar]
A.
B.
C.
D.
Q104
Objective
1. Taking it together
MCQ
A small block of mass m is kept on a rough inclined surface of inclination $\theta$ fixed in an elevator. The elevator goes up with a uniform velocity v and the block does not slide on the wedge.
The work done by the force of friction on the block in a time t will be
A.
zero
B.
$mgvt \cos^{2} \theta$
C.
$mgvt \sin^{2} \theta$
D.
$\frac{1}{2} mgvt \sin 2\theta$
Q105
Objective
1. Taking it together
MCQ
A pendulum of mass 1 kg and length l = 1m is released from rest at angle $\theta = 60^{\circ}$ . The power delivered by all the forces acting on the bob at angle $\theta = 30^{\circ}$ will be (Take, $g = 10 \, ms^{-2}$ )
A.
13.5 W
B.
20.4 W
C.
24.6 W
D.
zero
Q106
Objective
1. Taking it together
MCQ
A uniform flexible chain of mass m and length l hangs in equilibrium over a smooth horizontal pin of negligible diameter. One end of the chain is given a small vertical displacement, so that the chain slips over the pin. The speed of chain when it leaves pin is
A.
$\sqrt{\frac{gl}{2}}$
B.
$\sqrt{gl}$
C.
$\sqrt{2gl}$
D.
$\sqrt{3gl}$
Q107
Objective
1. Taking it together
MCQ
The potential energy of a particle of mass 1 kg is $U = 10 + (x - 2)^{2}$ . Here, U is in joule and x in metre on the positive X-axis. Particle travels up to $x = +6 \, m$ . Choose the correct option.
A.
On negative $X$ -axis, particle travels up to $x = -2\mathrm{m}$
B.
The maximum kinetic energy of the particle is 16 J
C.
Both a) and b) are correct
D.
Both a) and b) are incorrect
Q108
Objective
1. Taking it together
MCQ
A plank of mass 10 kg and a block of mass 2 kg are placed on a horizontal plane as shown in the figure. There is no friction between plane and plank. The coefficient of friction between block and plank is 0.5. A force of 60 N is applied on plank horizontally. In first 2 s, the work done by friction on the block is
A.
−100 J
B.
100 J
C.
zero
D.
200 j
Q109
Objective
1. Taking it together
MCQ
A block of mass m is pulled along a horizontal surface by applying a force at an angle $\theta$ with the horizontal. If the block travels with a uniform velocity and has a displacement d and the coefficient of friction is $\mu$ , then the work done by the applied force is
A.
$\frac{\mu mgd}{\cos\theta + \mu \sin\theta}$
B.
$\frac{\mu mgd\cos\theta}{\cos\theta + \mu\sin\theta}$
C.
$\frac{\mu mgd \sin\theta}{\cos\theta + \mu \sin\theta}$
D.
$\frac{\mu mgd\cos\theta}{\cos\theta - \mu\sin\theta}$
Q110
Objective
1. Taking it together
MCQ
An ideal massless spring S can be compressed 1 m by a force of 100 N in equilibrium. The same spring is placed at the bottom of a frictionless inclined plane inclined at $30^{\circ}$ to the horizontal. A 10 kg block M is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring by 2 m. If $g = 10 \, ms^{-2}$ , what is the speed of mass just before it touches the spring?
A.
$\sqrt{20}\ \mathrm{m\,s^{-1}}$
B.
$\sqrt{30}\ \mathrm{m\,s^{-1}}$
C.
$\sqrt{10}\ \mathrm{m\,s^{-1}}$
D.
$\sqrt{40}\ \mathrm{m\,s^{-1}}$
Q111
Objective
2. Assertion and reason
MCQ
Assertion At stable equilibrium position of a body, kinetic energy cannot be zero. Because it is maximum.
Reason During oscillations of a body, potential energy is minimum at stable equilibrium position.
A.
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
B.
If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
C.
If Assertion is correct but Reason is incorrect.
D.
If Assertion is incorrect but Reason is correct.
Q112
Objective
2. Assertion and reason
MCQ
Assertion If work done by conservative force is negative, then potential energy associated with that force should increase.
Reason This is according to the relation, $\Delta U = -W$ . Here, $\Delta U$ is change in potential energy and W is work done by conservative force.
A.
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
B.
If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
C.
If Assertion is correct but Reason is incorrect.
D.
If Assertion is incorrect but Reason is correct.
Q113
Objective
2. Assertion and reason
MCQ
Assertion If the surface between the blocks A and B is rough, then work done by friction on block B is always negative.
Reason Total work done by friction in both the blocks is always zero.
Reason Total work done by friction in both the blocks is always zero.
A.
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
B.
If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
C.
If Assertion is correct but Reason is incorrect.
D.
If Assertion is incorrect but Reason is correct.
Q114
Objective
2. Assertion and reason
MCQ
Assertion Force applied on a block moving in one dimension is producing a constant power, then the motion should be uniformly accelerated.
Reason This constant power multiplied with time is equal to the change in kinetic energy.
A.
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
B.
If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
C.
If Assertion is correct but Reason is incorrect.
D.
If Assertion is incorrect but Reason is correct.
Q115
Objective
2. Assertion and reason
MCQ
Assertion Total work done by spring may be positive, negative or zero.
Reason Direction of spring force is always towards mean position.
A.
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
B.
If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
C.
If Assertion is correct but Reason is incorrect.
D.
If Assertion is incorrect but Reason is correct.
Q116
Objective
2. Assertion and reason
MCQ
Assertion At any instant, the magnitude of rate of change of potential energy of the projectile of mass
1 kg is numerically equal to magnitude of $a \cdot v$ (where, a is acceleration due to gravity and v is velocity at that instant)
Reason The graph representing power delivered by the gravitational force acting on the projectile with time will be straight line with negative slope.
A.
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
B.
If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
C.
If Assertion is correct but Reason is incorrect.
D.
If Assertion is incorrect but Reason is correct.
Q117
Objective
3. Statement based questions
MCQ
Mark out the correct statement(s).
A.
Total work done by internal forces on a system is always zero.
B.
Total work done by internal forces on a system may sometimes be zero.
C.
Total work done by friction can never be zero.
D.
Total work done by friction is always zero.
Q118
Objective
3. Statement based questions
MCQ
Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track as shown in figure. [NCERT Exemplar]
Which of the following statement(s) is/are correct?
A.
Both the stones reach the bottom at the same time but not with the same speed.
B.
Both the stones reach the bottom with the same speed and stone I reaches the bottom earlier than stone II.
C.
Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I.
D.
Both the stones reach the bottom at different times and with different speeds.
Q119
Objective
3. Statement based questions
MCQ
In position A, kinetic energy of a particle is 60 J and potential energy is -20 J. In position B, kinetic energy is 100 J and potential energy is 40 J. Choose the correct statement, if the particle moves from A to B.
A.
Work done by conservative forces is 50 J.
B.
Work done by external forces is 40 J.
C.
Net work done by all the forces is 40 J.
D.
Net work done by all the forces is 100 J.
Q120
Objective
3. Statement based questions
MCQ
Which of the following statement(s) is/are correct?
I. Spring force is a conservative force.
II. Potential energy is defined only for conservative forces.
A.
Only I
B.
Only II
C.
Both I and II
D.
Neither I nor II
Q121
Objective
3. Statement based questions
MCQ
Which of the following statement(s) is/are correct?
I. If momentum of a body increases by 50%, its kinetic energy will increase by 125%.
II. Kinetic energy is proportional to square of velocity.
A.
Only I
B.
Only II
C.
Both I and II
D.
Neither I nor II
Q122
Objective
4. Match the columns
Numerical
A body is moved along a straight line by a machine delivering a power proportional to time $P \propto t$ . Then, match the following columns and mark the correct option from the codes given below.
| Column I | Column II | |||||||
| (A) | Velocity is proportional to | (p) | $t$ | |||||
| (B) | Displacement is proportional to | (q) | $t^{2}$ | |||||
| (C) | Work done is proportional to | (r) | $t^{3}$ | |||||
| Codes | ||||||||
| A | B | C | A | B | C | |||
| (a) | p | q | r | (b) | p | p | r | |
| (c) | p | q | q | (d) | r | p | q | |
Correct Answer: C
Explanation:
Power, $P \propto t$
Work done, $W = \int Pdt = \int \alpha t dt$ or $W \propto t^2$
Since, work done is equal to change in kinetic energy.
Hence,
$v ^ {2} \propto t ^ {2} \mathrm{or} v \propto t$
Further,
$v = \frac {d s}{d t}$
$\therefore \quad \frac{ds}{dt} \propto t \text{ or } ds \propto t dt \text{ or } s \propto t^2$
(By integration)
Hence, A → p, B → q, C → q.
Work done, $W = \int Pdt = \int \alpha t dt$ or $W \propto t^2$
Since, work done is equal to change in kinetic energy.
Hence,
$v ^ {2} \propto t ^ {2} \mathrm{or} v \propto t$
Further,
$v = \frac {d s}{d t}$
$\therefore \quad \frac{ds}{dt} \propto t \text{ or } ds \propto t dt \text{ or } s \propto t^2$
(By integration)
Hence, A → p, B → q, C → q.
Q123
Objective
4. Match the columns
Numerical
F-x and corresponding U-x graphs are as shown in figures. Three points A, B and C in F-x graph may be corresponding to P, Q and R in the U-x graph. Match the following columns and mark the correct option from the codes given below.
| Column I | Column II | |||
| (A) | A | (p) | P | |
| (B) | B | (q) | Q | |
| (C) | C | (r) | R | |
| (s) | None | |||
| Codes | ||||
| A | B | C | ||
| (a) | r | s | p | |
| (b) | p | s | r | |
| (c) | p | r | q | |
| (d) | q | r | s | |
Correct Answer: A
Explanation:
A is the point of stable equilibrium, so potential energy is minimum, i.e. at R. Similarly, point C is the unstable equilibrium position, where potential energy should be maximum, i.e. at P.
Hence, A → r, B → s, C → p.
Hence, A → r, B → s, C → p.
Q124
Objective
4. Match the columns
Numerical
Acceleration versus x and potential energy versus x graph of a particle moving along X-axis are as shown below. Mass of the particle is 1 kg and velocity at x = 0 is 4 ms $^{-1}$ .
Match the following columns for x = 8m and mark the correct option from the codes given below.
| Column I | Column II | ||||||||
| (A) | Final kinetic energy | (p) | 120 J | ||||||
| (B) | Work done by conservative forces | (q) | 240 J | ||||||
| (C) | Total work done | (r) | -120 J | ||||||
| (D) | Work done by external forces | (s) | 128 J | ||||||
| Codes | |||||||||
| A | B | C | D | A | B | C | D | ||
| (a) | p | s | r | q | (b) | s | q | p | r |
| (c) | s | p | r | q | (d) | p | q | r | s |
Correct Answer: B
Explanation:
$v_{f} - v_{i} = \text{Area of } a - x \text{ graph} = 12 \, \text{ms}^{-1}$
$\therefore v_{f} = 12 + 4 = 16\mathrm{ms}^{-1}$
$\Delta \mathrm{KE} = \frac {1}{2} m (v _ {f} ^ {2} - v _ {i} ^ {2}) = 1 2 0 \mathrm{J}$
Total work done by all the forces = $\Delta$ KE = 120 J
Final KE, $K_{f}=\frac{1}{2}mv_{f}^{2}=128J$
Work done by conservative forces = $U_{i} - U_{f} = 240$ J
Work done by external forces = Total work done - Work done by conservative forces = -120 J
$\text { Hence, } A \to s, B \to q, C \to p, D \to r.$
$\therefore v_{f} = 12 + 4 = 16\mathrm{ms}^{-1}$
$\Delta \mathrm{KE} = \frac {1}{2} m (v _ {f} ^ {2} - v _ {i} ^ {2}) = 1 2 0 \mathrm{J}$
Total work done by all the forces = $\Delta$ KE = 120 J
Final KE, $K_{f}=\frac{1}{2}mv_{f}^{2}=128J$
Work done by conservative forces = $U_{i} - U_{f} = 240$ J
Work done by external forces = Total work done - Work done by conservative forces = -120 J
$\text { Hence, } A \to s, B \to q, C \to p, D \to r.$
Q125
Objective
4. Match the columns
Numerical
A block of mass m is stationary with respect to a rough wedge is shown in figure. Starting from rest in time $t \left( m = 1 \text{kg}, \theta = 30^{\circ}, a = 2 \text{ms}^{-2}, t = 4 \text{s} \right)$ .
Match the following columns for work done on the block and mark the correct option from the codes given below.
| Column I | Column II | ||||
| (A) | By gravity | (p) | 144 J | ||
| (B) | By normal reaction | (q) | 32 J | ||
| (C) | By friction | (r) | -160 J | ||
| (D) | By all the forces | (s) | 48 J | ||
| Codes | B | C | D | ||
| (a) | r | q | s | p | |
| (b) | q | r | s | p | |
| (c) | p | q | r | s | |
| (d) | r | p | s | q | |
Correct Answer: D
Explanation:
In t = 4 s, v = at = 8 ms $^{-1}$ and s = $\frac{1}{2}$ at $^{2}$ = 16 m
$\mathrm{KE} = \frac {1}{2} m v ^ {2} = 3 2 \mathrm{J}$
From work-energy theorem,
Work done by all the forces = $\Delta KE = 32J$
Work done by gravity, $W_{g} = -mgh = -(1)(10)(16) = -160$ J
Writing equation of motion, we have
$N\cos 30^{\circ} + f\sin 30^{\circ} - 10 = ma = 2$
or
$\sqrt {3} N + f = 2 4$...(i)
$\Sigma F _ {x} = 0$
$N \sin 3 0 ^ {\circ} = f \cos 3 0 ^ {\circ} \mathrm{or} N = \sqrt {3} f$...(ii)
Solving Eqs. (i) and (ii), we have
$f = 6 \mathrm{N} \quad \text {and} \quad N = 6 \sqrt {3} \mathrm{N}$
Now, work done by normal reaction, $W_{N} = (N\cos \theta)(s)$
$= (6 \sqrt {3}) \left(\frac {\sqrt {3}}{2}\right) (1 6) = 1 4 4 \mathrm{J}$
Work done by friction, $W_{f} = (f\sin \theta)(s) = (6)\frac{1}{2}(16) = 48\mathrm{J}$
Work done by all the forces,
$W = W _ {g} + W _ {N} + W _ {F} = - 1 6 0 + 1 4 4 + 4 8 = 3 2 \mathrm{J}$
Hence, A → r, B → p, C → s, D → q.
$\mathrm{KE} = \frac {1}{2} m v ^ {2} = 3 2 \mathrm{J}$
From work-energy theorem,
Work done by all the forces = $\Delta KE = 32J$
Work done by gravity, $W_{g} = -mgh = -(1)(10)(16) = -160$ J
Writing equation of motion, we have
$N\cos 30^{\circ} + f\sin 30^{\circ} - 10 = ma = 2$
or
$\sqrt {3} N + f = 2 4$...(i)
$\Sigma F _ {x} = 0$
$N \sin 3 0 ^ {\circ} = f \cos 3 0 ^ {\circ} \mathrm{or} N = \sqrt {3} f$...(ii)
Solving Eqs. (i) and (ii), we have
$f = 6 \mathrm{N} \quad \text {and} \quad N = 6 \sqrt {3} \mathrm{N}$
Now, work done by normal reaction, $W_{N} = (N\cos \theta)(s)$
$= (6 \sqrt {3}) \left(\frac {\sqrt {3}}{2}\right) (1 6) = 1 4 4 \mathrm{J}$
Work done by friction, $W_{f} = (f\sin \theta)(s) = (6)\frac{1}{2}(16) = 48\mathrm{J}$
Work done by all the forces,
$W = W _ {g} + W _ {N} + W _ {F} = - 1 6 0 + 1 4 4 + 4 8 = 3 2 \mathrm{J}$
Hence, A → r, B → p, C → s, D → q.