Work, Power & Energy
304 Questions
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Online September 2020
Q1
Advance
7. ARCHIVE: JEE MAIN
MCQ
A person pushes a box on a rough horizontal platform surface. He applies a force of 200 N over a distance of 15 m. Thereafter, he gets progressively tired and his applied force reduces linearly with distance to 100 N. The total distance through which the box has been moved is 30 m. What is the work done by the person during the total movement of the box?
A.
5690 J
B.
5250 J
C.
3280 J
D.
2780 J
Online September 2020
Q2
Advance
7. ARCHIVE: JEE MAIN
MCQ
If the potential energy between two molecules is given by $U = -\frac{A}{r^6} + \frac{B}{r^{12}}$ , then at equilibrium, separation between molecules, and the potential energy are
A.
$\left(\frac{B}{A}\right)^{\frac{1}{6}}, 0$
B.
$\left(\frac{B}{2A}\right)^{\frac{1}{6}}, -\frac{A^2}{2B}$
C.
$\left(\frac{2B}{A}\right)^{\frac{1}{6}}, -\frac{A^2}{4B}$
D.
$\left(\frac{2B}{A}\right)^{\frac{1}{6}}, -\frac{A^2}{2B}$
Online September 2020
Q3
Advance
7. ARCHIVE: JEE MAIN
MCQ
A particle is moving unidirectionally on a horizontal plane under the action of a constant power supplying energy source. The displacement $(s)$ time $(t)$ graph that describes the motion of the particle is (Graphs are drawn schematically and are not to scale)
A.
B.
C.
D.
Online September 2020
Q4
Advance
7. ARCHIVE: JEE MAIN
Numerical
A body of mass 2 kg is driven by an engine delivering a constant power $1 \, Js^{-1}$ . The body starts from rest and moves in a straight line. After 9 second, the body has moved a distance ..... (in metre).
Correct Answer: 18
Explanation:
Since, $P =$ constant, so from Work-Energy Theorem,
$\mathrm{weget} W = \frac {1}{2} m v ^ {2}. \mathrm{Also}, W = P t$
$\begin{array}{r l} \Rightarrow & K = P t = \frac {1}{2} m v ^ {2} \mathrm{i.e.,} v = \sqrt {\frac {2 P t}{m}} \\ \Rightarrow & v = \frac {d x}{d t} = \sqrt {\frac {2 P t}{m}} = \sqrt {\frac {2 P}{m}} t ^ {1 / 2} \\ \Rightarrow & x = \int d x = \sqrt {\frac {2 P}{m}} \int_ {0} ^ {t} t ^ {1 / 2} d t = \frac {2}{3} \sqrt {\frac {2 P}{m}} t ^ {3 / 2} \\ \Rightarrow & x = \frac {2}{3} \sqrt {\frac {2 (1)}{2}} (9) ^ {3 / 2} = \frac {2}{3} (1) (2 7) = 1 8 \mathrm{m} \end{array}$
Hence, the correct answer is (18).
Online September 2020
Q5
Advance
7. ARCHIVE: JEE MAIN
Numerical
A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine so that it rises to a maximum height of 20 m after leaving the machine. If the part pushing the ball applies a constant force F on the ball and moves horizontally a distance of 0.2 m while launching the ball, the value of F (in N) is ..... $g = 10 \text{ ms}^{-2}$
Correct Answer: 150
Explanation:
$\begin{array}{r l} & {\mathrm{Since,} W _ {F} = \frac {1}{2} m v ^ {2} = m g h} \\ & {\Rightarrow F s = m g h} \\ & {\Rightarrow F (0. 2) = (0. 1 5) (1 0) (2 0)} \\ & {\Rightarrow F = 1 5 0 \mathrm{N}} \end{array}$
Hence, the correct answer is (150).
Hence, the correct answer is (150).
Online September 2020
Q6
Advance
7. ARCHIVE: JEE MAIN
Numerical
A small block starts slipping down from a point B on an inclined plane AB, which is making an angle $\theta$ with the horizontal section BC is smooth and the remaining section CA is rough with a coefficient of friction $\mu$ . It is found that the block comes to rest as it reaches the bottom (point A) of the inclined plane. If BC = 2AC, the coefficient of friction is given by $\mu = k \tan \theta$ . The value of k is .....
Correct Answer: 3
Explanation:
Applying Work-Energy Theorem, we get
$m g \sin \theta (A C + 2 A C) - (\mu m g \cos \theta) A C = 0$
$\Rightarrow \quad \mu = 3 \tan \theta$
$\Rightarrow k = 3$
Hence, the correct answer is (3).
$m g \sin \theta (A C + 2 A C) - (\mu m g \cos \theta) A C = 0$
$\Rightarrow \quad \mu = 3 \tan \theta$
$\Rightarrow k = 3$
Hence, the correct answer is (3).
Online January 2020
Q7
Advance
7. ARCHIVE: JEE MAIN
MCQ
An elevator in a building can carry a maximum of 10 persons, with the average mass of each person being 68 kg. The mass of the elevator itself is 920 kg and it moves with a constant speed $3 \, ms^{-1}$ . The frictional force opposing the motion is 6000 N. If the elevator is moving up with its full capacity, the power delivered by the motor to the elevator ( $g = 10 \, ms^{-2}$ ) must be at least
A.
56300 W
B.
48000 W
C.
66000 W
D.
62360 W
Online January 2020
Q8
Advance
7. ARCHIVE: JEE MAIN
MCQ
Consider a force $\vec{F} = -x\hat{i} + y\hat{j}$ . The work done by this force in moving a particle from point $A(1,0)$ to $B(0,1)$ along the line segment shown is .... (All quantities are in SI units)
A.
$\frac{3}{2}$
B.
1
C.
2
D.
$\frac{1}{2}$
Online January 2020
Q9
Advance
7. ARCHIVE: JEE MAIN
Numerical
A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to .... ms $^{-1}$ . Take 1 HP = 746 W and g = 10 ms $^{-2}$ .
Correct Answer: 1.9
Explanation:
Let the elevator be moving up with a constant speed v. Then frictional force f will be downwards. So, tension in the cable is
T = 2 0 0 0 g + f = 2 0 0 0 + 4 0 0 0
$\Rightarrow T = 2 4 0 0 0 \mathrm{N}$
Power of the electric motor of the elevator is
P = F v = T v
$\Rightarrow 6 0 \times 7 4 6 = (2 4 0 0 0) v$
$\Rightarrow v = \frac {6 0 \times 7 4 6}{2 4 0 0 0} = 1. 8 6 5 \approx 1. 9 \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (1.9).
T = 2 0 0 0 g + f = 2 0 0 0 + 4 0 0 0
$\Rightarrow T = 2 4 0 0 0 \mathrm{N}$
Power of the electric motor of the elevator is
P = F v = T v
$\Rightarrow 6 0 \times 7 4 6 = (2 4 0 0 0) v$
$\Rightarrow v = \frac {6 0 \times 7 4 6}{2 4 0 0 0} = 1. 8 6 5 \approx 1. 9 \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (1.9).
Online January 2020
Q10
Advance
7. ARCHIVE: JEE MAIN
Numerical
A particle (m = 1 kg) slides down a frictionless track (AOC) starting from rest at a point A (height 2 m). After reaching C, the particle continues to move freely in air as a projectile. When it reaches its highest point P (height 1 m), the kinetic energy of the particle (in J) is ..... (Figure drawn is schematic and not to scale, take $g = 10 \, ms^{-2}$ )
Correct Answer: 10
Explanation:
Applying Law of Conservation of Mechanical Energy between A and B, we get
$\begin{array}{l l} & U _ {A} + K _ {A} = U _ {B} + K _ {B} \\ \Rightarrow & m g (2) = m g (1) + K _ {2} \\ \Rightarrow & K _ {2} = m g = 1 0 \mathrm{J} \\ \text { Hence,the correct answer is(10). } \end{array}$
$\begin{array}{l l} & U _ {A} + K _ {A} = U _ {B} + K _ {B} \\ \Rightarrow & m g (2) = m g (1) + K _ {2} \\ \Rightarrow & K _ {2} = m g = 1 0 \mathrm{J} \\ \text { Hence,the correct answer is(10). } \end{array}$
2020
Q11
Advance
8. ARCHIVE: JEE ADVANCED
MSQ
JEE (Advanced) 2020
A student skates up a ramp makes an angle $30^{\circ}$ with the horizontal. He/she starts (as shown in Figure) at the bottom of the ramp with speed $v_{0}$ and wants to turn around over a semicircular path xyz of radius R during which he/she reaches a maximum height h (at point y) from the ground as shown in Figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then (g is the acceleration due to gravity)
A.
$v_{0}^{2}-2gh=\frac{1}{2}gR$
B.
$v_{0}^{2}-2gh=\frac{\sqrt{3}}{2}gR$
C.
the centripetal force required at points x and z is zero
D.
the centripetal force required is maximum at points $x$ and $z$
Online April 2019
Q12
Advance
7. ARCHIVE: JEE MAIN
MCQ
A particle moves in one dimension from rest under the influence of a force that varies with the distance travelled by the particle as shown in the Figure. The kinetic energy of the particle after it has travelled 3 m is
A.
4 J
B.
2.5 J
C.
5 J
D.
6.5 J
Online April 2019
Q13
Advance
7. ARCHIVE: JEE MAIN
MCQ
A uniform cable of mass M and length L is placed on a horizontal surface such that its $\left(\frac{1}{n}\right)^{\text{th}}$ part is hanging below the edge of the surface. To lift the hanging part of the cable up to the surface, the work done should be
A.
$\frac{MgL}{n^{2}}$
B.
nMgL
C.
$\frac{MgL}{2n^{2}}$
D.
$\frac{2MgL}{n^{2}}$
Online January 2019
Q14
Advance
7. ARCHIVE: JEE MAIN
MCQ
A block of mass $m$ , lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant $k$ . The other end of the spring is fixed, as shown in the figure. The block is initially at rest in its equilibrium position. If now the block is pulled with a constant force $F$ , the maximum speed of the block is
A.
$\frac{F}{\pi\sqrt{mk}}$
B.
$\frac{\pi F}{\sqrt{mk}}$
C.
$\frac{2F}{\sqrt{mk}}$
D.
$\frac{F}{\sqrt{mk}}$
Online January 2019
Q15
Advance
7. ARCHIVE: JEE MAIN
MCQ
A force acts on a 2 kg object so that its position is given as a function of time as $x = 3t^{2} + 5$ . What is the work done by this force in first 5 seconds?
A.
950 J
B.
900 J
C.
850 J
D.
875 J
Online January 2019
Q16
Advance
7. ARCHIVE: JEE MAIN
MCQ
A block of mass $m$ is kept on a platform which starts from rest with constant acceleration $\frac{g}{2}$ upward, as shown in figure. Work done by normal reaction on block in time $t$ is
A.
$\frac{3mg^2t^2}{8}$
B.
$-\frac{mg^2t^2}{8}$
C.
0
D.
$\frac{mg^2t^2}{8}$
Online January 2019
Q17
Advance
7. ARCHIVE: JEE MAIN
MCQ
A particle which is experiencing a force, given by $\vec{F} = 3\hat{i} - 12\hat{j}$ , undergoes a displacement of $\vec{d} = 4\hat{i}$ . If the particle had a kinetic energy of $3\mathrm{J}$ at the beginning of the displacement, what is its kinetic energy of the end of the displacement?
A.
15 J
B.
9 J
C.
12 J
D.
10 J
Online January 2019
Q18
Advance
7. ARCHIVE: JEE MAIN
MCQ
A pendulum is executing simple harmonic motion and its maximum kinetic energy is $K_{1}$ . If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is $K_{2}$ . Then
A.
$K_{2} = 2K_{1}$
B.
$K_{2} = \frac{K_{1}}{4}$
C.
$K_{2} = K_{1}$
D.
$K_{2} = \frac{K_{1}}{2}$
JEE (Advanced) 2019
Q19
Advance
8. ARCHIVE: JEE ADVANCED
Numerical
Aparticle is moved along a path $AB-BC-CD-DE-EF-FA$ , as shown in figure, in presence of a force $\vec{F} = (\alpha y\hat{i} + 2\alpha x\hat{j})$ N, where $x$ and $y$ are in meter and $\alpha = -1$ Nm $^{-1}$ . The work done on the particle by this force $\vec{F}$ will be____ joule.
Correct Answer: 0.75
Explanation:
Given that, $\vec{F} = \alpha y \hat{i} + 2 \alpha x \hat{j}$ where $\alpha = -1$
Since Work done is given by
$\begin{array}{r l} & W = \int \vec {F} \cdot d \vec {\ell} \\ \Rightarrow & d W = \alpha (y d x + 2 x d y) \end{array}$
For $A\to B$ , $y = 1$ , so $dy = 0$
$\Rightarrow W _ {A \rightarrow B} = \alpha \int y d x = \alpha \int_ {0} ^ {1} d x = \alpha$
For $B\to C$ , $x = 0$ , so $dx = 0$
$\Rightarrow W _ {B \rightarrow C} = 2 \alpha \int_ {1} ^ {0. 5} d y = 2 \alpha (- 0. 5) = - \alpha$
For $C\to D$ , $y = 0.5$ , so $dy = 0$
$\Rightarrow W _ {C \rightarrow D} = 0. 5 \alpha \int_ {1} ^ {0.5} d x = - \frac {\alpha}{4}$
For $D\to E$ , $x = 0.5$ , so $dx = 0$
$\Rightarrow W _ {D \rightarrow E} = 2 \alpha (0.5) \int_ {0.5} ^ {0} d y = - \frac {\alpha}{2}$
$\Rightarrow W _ {E \to F} = 0$
$\Rightarrow W _ {F \to A} = 0$
Hence total work done is
$\begin{array}{r l} & W _ {t o t a l} = \alpha - \alpha - \frac {\alpha}{4} - \frac {\alpha}{2} \\ \Rightarrow & W _ {t o t a l} = - \frac {3 \alpha}{4} = \frac {- 3 (- 1)}{4} = \frac {3}{4} \mathrm{J} = 0. 7 5 \mathrm{J} \end{array}$
Since Work done is given by
$\begin{array}{r l} & W = \int \vec {F} \cdot d \vec {\ell} \\ \Rightarrow & d W = \alpha (y d x + 2 x d y) \end{array}$
For $A\to B$ , $y = 1$ , so $dy = 0$
$\Rightarrow W _ {A \rightarrow B} = \alpha \int y d x = \alpha \int_ {0} ^ {1} d x = \alpha$
For $B\to C$ , $x = 0$ , so $dx = 0$
$\Rightarrow W _ {B \rightarrow C} = 2 \alpha \int_ {1} ^ {0. 5} d y = 2 \alpha (- 0. 5) = - \alpha$
For $C\to D$ , $y = 0.5$ , so $dy = 0$
$\Rightarrow W _ {C \rightarrow D} = 0. 5 \alpha \int_ {1} ^ {0.5} d x = - \frac {\alpha}{4}$
For $D\to E$ , $x = 0.5$ , so $dx = 0$
$\Rightarrow W _ {D \rightarrow E} = 2 \alpha (0.5) \int_ {0.5} ^ {0} d y = - \frac {\alpha}{2}$
$\Rightarrow W _ {E \to F} = 0$
$\Rightarrow W _ {F \to A} = 0$
Hence total work done is
$\begin{array}{r l} & W _ {t o t a l} = \alpha - \alpha - \frac {\alpha}{4} - \frac {\alpha}{2} \\ \Rightarrow & W _ {t o t a l} = - \frac {3 \alpha}{4} = \frac {- 3 (- 1)}{4} = \frac {3}{4} \mathrm{J} = 0. 7 5 \mathrm{J} \end{array}$
2018
Q20
Advance
7. ARCHIVE: JEE MAIN
MCQ
A particle is moving in a circular path of radius $a$ under the action of an attractive potential $U = -\frac{k}{2r^2}$ . Its total energy is
A.
$-\frac{k}{4a^2}$
B.
$\frac{k}{2a^2}$
C.
Zero
D.
$-\frac{3}{2}\frac{k}{a^2}$
Online 2018
Q21
Advance
7. ARCHIVE: JEE MAIN
MCQ
Two particles of the same mass m are moving in circular orbits because of force, given by $F(r) = -\frac{16}{r} - r^{3}$ . The first particle is at a distance r = 1 and the second, at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to
A.
$10^{-1}$
B.
$6 \times 10^{-2}$
C.
$6 \times 10^{2}$
D.
$3 \times 10^{-3}$
Online 2018
Q22
Advance
7. ARCHIVE: JEE MAIN
MCQ
A body of mass m starts moving from rest along x-axis so that its velocity varies as $v = a\sqrt{s}$ where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first t seconds after the start of the motion is
A.
$\frac{1}{4}ma^{4}t^{2}$
B.
$4ma^{4}t^{2}$
C.
$\frac{1}{8}ma^{4}t^{2}$
D.
$8ma^{4}t^{2}$
2017
Q23
Advance
7. ARCHIVE: JEE MAIN
MCQ
A body of mass $m=10^{-2}$ kg is moving in a medium and experiences a frictional force $F=-kv^{2}$ . Its initial speed is $v_{0}=10~ms^{-1}$ . If, after 10 s, its energy is $\frac{1}{8}mv_{0}^{2}$ , the value of k will be
A.
$10^{-3}$ $kgm^{-1}$
B.
$10^{-3}$ $kgs^{-1}$
C.
$10^{-4}$ $kgm^{-1}$
D.
$10^{-1}$ $kgm^{-1}s^{-1}$
2017
Q24
Advance
7. ARCHIVE: JEE MAIN
MCQ
A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 second will be
A.
4.5 J
B.
22 J
C.
9 J
D.
18 J
2016
Q25
Advance
7. ARCHIVE: JEE MAIN
MCQ
A point particle having some mass, moves along the uniformly rough track $PQR$ as shown in the figure. The coefficient of friction, between the particle and the rough track equals $\mu$ . The particle is released, from rest, from the point $P$ and it comes to rest at a point $R$ . The energies, lost by the ball, over the parts, $PQ$ and $QR$ , of the track, are equal to each other, and no energy is lost when particle changes direction from $PQ$ to $QR$ . The values of the coefficient of friction $\mu$ and the distance $x(= QR)$ , are respectively close to
A.
0.29 and $6.5\mathrm{m}$
B.
0.2 and $6.5\mathrm{m}$
C.
0.2 and $3.5\mathrm{m}$
D.
0.29 and $3.5\mathrm{m}$
