Work, Power & Energy
227 Questions
Start Allen Test
Q51
Allen
2. Dynamics of Circular Motion
MCQ
A long horizontal rod has a bead which can slide along its length and is initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with a constant angular acceleration, $\alpha$ . If the coefficient of friction between the rod and bead is $\mu$ , and gravity is neglected, then the time after which the bead starts slipping is:-
A.
$\sqrt{\frac{\mu}{\alpha}}$
B.
$\frac{\mu}{\sqrt{\alpha}}$
C.
$\frac{1}{\sqrt{\mu\alpha}}$
D.
Infinitesimal
Q52
Allen
2. Dynamics of Circular Motion
MCQ
A car is moving along a circular track with tangential acceleration of magnitude $a_{0}$ . It just start to slip at speed $v_{0}$ then find radius of circle. (Coefficient of friction is $\mu$ )
A.
$\frac{v_{0}^{2}}{\sqrt{(\mu g)^{2} + a_{0}^{2}}}$
B.
$\frac{v_{0}^{2}}{\sqrt{(\mu g)^{2} - a_{0}^{2}}}$
C.
$\frac{v_{0}^{2}}{\sqrt{(\mu g)^{2} - 2a_{0}^{2}}}$
D.
$\frac{v_{0}^{2}}{\sqrt{(\mu g)^{2} + 2a_{0}^{2}}}$
Q53
Allen
2. Dynamics of Circular Motion
MCQ
A bead is constrained to move on rod in gravity free space as shown in figure. The rod is rotating with angular velocity $\omega$ and angular acceleration $\alpha$ about its end. If $\mu$ is coefficient of friction. Mark the correct option. (Rod rotates in the plane of paper)
A.
If $\mu = \frac{\omega^2}{\alpha}$ friction on bead is static in nature
B.
If $\mu > \frac{\omega^2}{\alpha}$ friction on bead is kinetic in nature
C.
If $\mu < \frac{\omega^2}{\alpha}$ friction is static
D.
If bead does not slide relative to rod. Friction will not exist between bead and rod.
Q54
Allen
2. Dynamics of Circular Motion
MCQ
A horizontal turn table is initially at rest. It starts to rotate in horizontal plane with constant angular acceleration $3 \, rad/s^{2}$ . A block of mass $1 \, kg$ is placed 1 metre away from the axis of rotation on the turn table. A horizontal force $F = 5t \, N$ where t is in seconds, is acting on the block always towards the center of rotation. The block begins to slip at t = 1s. The coefficient of friction between block and turn table is
A.
0.2
B.
0.3
C.
0.4
D.
0.5
Q55
Allen
3. Work
MCQ
A cricket player throws a ball straight upward in the air. It rises and falls back to the hands of the player. During the catch, his hands are displaced downward a few inches as the ball slows down, then.
A.
During the catch while the ball is in hand and is moving downward, the work done by the player on the ball is positive.
B.
During the throw work done by gravity is positive.
C.
During the throw while ball is in hand and moving upward, work done by ball on hand is negative.
D.
During the catch while the ball is in hand and moving downward, work done by hand plus gravity on the ball is positive.
Q56
Allen
3. Work
MCQ
A force of magnitude of 30 N acting along $\hat{i} + \hat{j} + \hat{k}$ , displaces a particle from point (2, 4, 1) m to (3, 5, 2) m. The work done during this displacement is :-
A.
90 J
B.
30 J
C.
$30\sqrt{3}J$
D.
$30/\sqrt{3}J$ .
Q57
Allen
3. Work
MCQ
A rope is used to lower vertically a block of mass M by a distance x with a constant downward acceleration g/2. The work done by the rope on the block is:
A.
Mgx
B.
$\frac{1}{2}$ Mgx $^{2}$
C.
$-\frac{1}{2}$ Mgx
D.
Mgx $^{2}$
Q58
Allen
3. Work
MCQ
A block of mass 10 kg is released from the top of smooth wedge (whose inclination at θ = 37°) which is moving with a constant velocity 10 m/s. The work done by the force applied by wedge on the block in 2 sec will be (in joules) :-
A.
1600
B.
960
C.
Zero
D.
720
Q59
Allen
3. Work
MCQ
In the figure shown find work done by tension in string on 3 kg block in first two second, from beginning of motion. The system is set free at t = 0 ( $g = 10 \, m/s^{2}$ ) (Assume pulley and string is ideal)
A.
-96 Joule
B.
-48 Joule
C.
+24 Joule
D.
96 Joule
Q60
Allen
3. Work
MCQ
Force acting on a particle is $(2\hat{i} + 3\hat{j})$ N. Work done by this force is zero, when a particle is moved on the line $3y + kx = 5$ . Here value of k is
A.
2
B.
4
C.
6
D.
8
Q61
Allen
3. Work
MCQ
A particle is moved from $(0,0)$ to $(a,a)$ under a force from $\vec{F}=\left(3\hat{i}+4\hat{j}\right)$ two paths. Path 1 is OP and path 2 is OQP. Let $W_{1}$ and $W_{2}$ be the work done by this force in these two paths. Then:
A.
$W_{1} = W_{2}$
B.
$W _ {1} = 2 W _ {2}$
C.
$W _ {2} = 2 W _ {1}$
D.
$W_{2} = 4W_{1}$
Q62
Allen
3. Work
MCQ
The components of a force acting on a particle are varying according to the graphs shown. To reach at point B (8, 20, 0) from point A(0, 5, 12) the particle moves on paths parallel to x-axis then y-axis and then z-axis, then work done by this force is :-
A.
$192J$
B.
$58J$
C.
250 J
D.
125 J
Q63
Allen
3. Work
MCQ
The work done by the force $\vec{F} = x^{2}\hat{i} + y^{2}\hat{j}$ around the path shown in the figure is
A.
$\frac{2}{3} a^{3}$
B.
zero
C.
$a^{3}$
D.
$\frac{4}{3} a^{3}$
Q64
Allen
3. Work
MCQ
By applying a force $\vec{F}=(3xy-5z)\hat{j}-4\hat{k}$ a particle is moved along the path $y=x^{2}$ from point $(0,0,0)$ to the point $(2,4,0)$ . The work done by the F on the particle is :-
A.
$\frac{280}{5}$
B.
$\frac{140}{5}$
C.
$\frac{232}{5}$
D.
$\frac{192}{5}$
Q65
Allen
4. Kinetic Energy
MCQ
A man who is running has half the kinetic energy of the boy of half his mass. The man speeds up by 1 m/s and then has the same kinetic energy as the boy. The original speed of the man was
A.
$\sqrt{2}$ m/s
B.
$(\sqrt{2}-1)$ m/s
C.
2 m/s
D.
$(\sqrt{2}+1)$ m/s
Q66
Allen
5. Work Energy Theorem
MCQ
A body of mass 0.5 kg travels in a straight line with velocity $v = ax^{3/2}$ where $a = 5 m^{-1/2}s^{-1}$ . The work done by the net force during its displacement from x = 0 to x = 2 m is
A.
1.5 J
B.
50 J
C.
10 J
D.
100 J
Q67
Allen
5. Work Energy Theorem
MCQ
In a shotput event an athlete throws the shotput of mass 10 kg with an initial speed of 1m/s at $45^{\circ}$ from a height 1.5 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be $10\ m/s^{2}$ , the kinetic energy of the shotput when it just reaches the ground will be
A.
2.5 J
B.
5.0 J
C.
52.5 J
D.
155.0 J
Q68
Allen
5. Work Energy Theorem
MCQ
In the figure, a block slides along a track from one level to a higher level, by moving through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed $v_{0}$ is 6 m/s, the height difference h is 1.1m and the coefficient of kinetic friction $\mu$ is 0.6. The value of d is
A.
1.17 m
B.
1.71 m
C.
7.11 m
D.
11.7 m
Q69
Allen
5. Work Energy Theorem
MCQ
A small object with mass $m = 2 \, kg$ is attached to the free end of an ideal spring with $k = 10 \, Nm^{-1}$ . The other end of the spring is connected to a fixed frictionless pivot located at the origin $O$ as shown in the Figure. The relaxed length of the spring is $1 \, m$ . An unspecified force $F$ carries the object at rest from point $A$ to point $B$ . At point $B$ , the object has the speed $5 \, ms^{-1}$ . The work done by the force $F$ is (Neglect gravity):-
A.
$130 \, J$
B.
$80 \, J$
C.
$-105 \, J$
D.
$-80 \, J$
Q70
Allen
5. Work Energy Theorem
MCQ
In the arrangement shown a block of mass 2kg is pressed against a spring which is compressed by 10 cm and is applying a force of 80 N to the block. The spring is not attached to the block. The coefficient of friction is 0.5. When the block is released it moves up the plane by some distance and then comes back down. The kinetic energy of the block just before it starts compressing the spring again is
A.
$2J$
B.
$1.6J$
C.
$0.8J$
D.
$0.4J$
Q71
Allen
5. Work Energy Theorem
MCQ
A block of mass 10kg is attached to a relaxed spring. Other end is attached to ceiling and block is released keeping the spring in vertical position. The spring is such that it has elongation of 10 cm under a force of 100 N. The maximum velocity of block in subsequent motion is:-
A.
1 m/s
B.
1.5 m/s
C.
2.0 m/s
D.
2.5 m/s
Q72
Allen
5. Work Energy Theorem
MCQ
A light spring of length 20 cm and force constant 2 N/cm is placed vertically on a table. A small block of mass 1 kg falls on it. The length h from the surface of the table at which the block will have the maximum velocity is:
A.
20 cm
B.
15 cm
C.
10 cm
D.
5cm
Q73
Allen
5. Work Energy Theorem
MCQ
In the figure shown, the system is released from rest. Find the velocity of block A when block B has fallen a distance ' $\ell'$ . Assume all pulleys to be massless and frictionless.
A.
$\sqrt{\frac{g\ell}{5}}$
B.
$\sqrt{g\ell}$
C.
$\sqrt{5g\ell}$
D.
None of these
Q74
Allen
5. Work Energy Theorem
Match the Columns
Two blocks, of masses M and 2M, are connected to a light spring of spring constant K that has one end fixed, as shown in figure. The horizontal surface and the pulley are frictionless. The blocks are released from rest when the spring is non deformed. The string is light.


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Correct Answer: B
Explanation:
Ans. (B)
Maximum extension will be at the moment when both masses stop momentarily after going down. Applying W-E theorem from starting to that instant.
$k _ {f} - k _ {\mathrm{i}} = W _ {g r.} + W _ {\mathrm{sp}} + \left(- \frac {1}{2} K x ^ {2}\right) W _ {t e n}.$
$0 - 0 = 2 \mathrm{M.g.x} + \left(- \frac {1}{2} K x ^ {2}\right) + 0$
$x = \frac {4 M g}{K}$
System will have maximum KE when net force on the system becomes zero. Therefore
$2 M g = T \mathrm{and} T = k x$
$\Rightarrow x = \frac {2 M g}{K}$
Hence $KE$ will be maximum when $2M$ mass has gone down by $\frac{2Mg}{K}$ .
Applying $W / E$ theorem
$k _ {f} - 0 = 2 M g. \frac {2 M g}{K} - \frac {1}{2} K. \frac {4 M ^ {2} g ^ {2}}{K ^ {2}}$
$k _ {f} = \frac {2 M ^ {2} g ^ {2}}{K ^ {2}}$
Maximum extension will be at the moment when both masses stop momentarily after going down. Applying W-E theorem from starting to that instant.
$k _ {f} - k _ {\mathrm{i}} = W _ {g r.} + W _ {\mathrm{sp}} + \left(- \frac {1}{2} K x ^ {2}\right) W _ {t e n}.$
$0 - 0 = 2 \mathrm{M.g.x} + \left(- \frac {1}{2} K x ^ {2}\right) + 0$
$x = \frac {4 M g}{K}$
System will have maximum KE when net force on the system becomes zero. Therefore
$2 M g = T \mathrm{and} T = k x$
$\Rightarrow x = \frac {2 M g}{K}$
Hence $KE$ will be maximum when $2M$ mass has gone down by $\frac{2Mg}{K}$ .
Applying $W / E$ theorem
$k _ {f} - 0 = 2 M g. \frac {2 M g}{K} - \frac {1}{2} K. \frac {4 M ^ {2} g ^ {2}}{K ^ {2}}$
$k _ {f} = \frac {2 M ^ {2} g ^ {2}}{K ^ {2}}$
Q75
Allen
6. Potential Energy
MCQ
A radial force $\vec{F} = \frac{A}{r^2}\hat{r}$ acts on a particle. The work done by the force in displacing a particle from (1, 0) to (3, 4) is:
A.
$\frac{A}{5}$
B.
$\frac{A}{2}$
C.
$\frac{4A}{5}$
D.
$\frac{4A}{3}$






