Work Energy and Power | Physics JEE previous year questions with Solutions

Q 11 :

Two bodies are having kinetic energies in the ratio 16 : 9. If they have same linear momentum, the ratio of their masses respectively is [2023]

4 : 3
3 : 4
16 : 9
9 : 16

1

2

3

4

(4)

$\frac{K_{1}}{K_{2}} = \frac{p_{1}^{2}}{2 m_{1}} \times \frac{2 m_{2}}{p_{2}^{2}} = \frac{m_{2}}{m_{1}} = \frac{16}{9}$

$\frac{m_{1}}{m_{2}} = \frac{9}{16}$

Q 12 :

An object of mass $' m'$ initially at rest on a smooth horizontal plane starts moving under the action of a force $F = 2 N$ . In the process of its linear motion, the angle $θ$ (as shown in the figure) between the direction of force and the horizontal varies as $θ = k x,$ where $' k'$ is a constant and $' x'$ is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be $E = \frac{n}{k} \sin θ .$ The value of $' n'$ is ________ . [2023]

[IMAGE 38]

0%

(2)

[IMAGE 39]

$θ = k x$

$W = K E_{f} - K E_{i}$

$d W = F d x$

$W = \int F_{1} d x = \int F \cos k x d x$

$= \frac{2}{k} (\sin k x)$

$∴$ $E = W = \frac{2}{k} \sin k x = \frac{2}{k} \sin θ$

Q 13 :

A lift of mass M = 500 kg is descending with speed of 2 m $s^{- 1}$ . Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of 2 m $s^{- 2}$ . The kinetic energy of the lift at the end of fall through a distance of 6 m will be ________ kJ. [2023]

0%

(7)

$v^{2} = u^{2} + 2 a s = 2^{2} + 2 (2) (6)$

$= 4 + 24 = 28$

$K E = \frac{1}{2} m v^{2} = \frac{1}{2} (500) (28) = 7000 J = 7 kJ$

Q 14 :

A particle of mass 10 g moves in a straight line with retardation $2 x$ , where $x$ is the displacement in SI units. Its loss of kinetic energy for above displacement is ${(\frac{10}{x})}^{- n} J .$ The value of $n$ will be _________ . [2023]

0%

(2)

$Loss of K.E. = work done against retarding force$

$= \int_{0}^{x} m a d x = \int_{0}^{x} m 2 x d x = m x^{2}$

$= (10^{- 2} kg) x^{2} J = {(\frac{10}{x})}^{- 2} J$

$So n = 2$

Q 15 :

A body is dropped on ground from a height $' h_{1}'$ and after hitting the ground, it rebounds to a height $' h_{2}'$ . If the ratio of velocities of the body just before and after hitting ground is 4, then percentage loss in kinetic energy of the body is $\frac{x}{4}$ . The value of $x$ is _________. [2023]

0%

(375)

$Let V_{1} and V_{2} be velocities just before and just after hitting the floor.$

$\frac{V_{1}}{V_{2}} = 4 \Rightarrow V_{1} = 4 V_{2}$

$K E_{before} = \frac{1}{2} m V_{1}^{2}$

$K E_{after} = \frac{1}{2} m V_{2}^{2} = \frac{1}{2} \frac{m V_{1}^{2}}{16}$

$Δ K E = \frac{1}{2} m V_{1}^{2} = (\frac{1}{16} - 1) = \frac{- 15}{32} m V_{1}^{2}$

$% change = \frac{Δ K E}{K E_{before}} \times 100 % = \frac{- 15}{16} \times 100 % = \frac{- 375}{4} %$

Q 16 :

The momentum of a body is increased by 50%. The percentage increase in the kinetic energy of the body is ________ %. [2023]

0%

(125)

$Kinetic energy of body = \frac{p^{2}}{2 m}$

$Initial kinetic energy = \frac{p_{i}^{2}}{2 m}$

$Final kinetic energy = \frac{p_{f}^{2}}{2 m} = \frac{{(1.5 p_{i})}^{2}}{2 m} = \frac{2.25 p_{i}^{2}}{2 m}$

$% increase in KE = \frac{2.25 \frac{p_{i}^{2}}{2 m} - \frac{p_{i}^{2}}{2 m}}{\frac{p_{i}^{2}}{2 m}} \times 100 = 125 %$

Q 17 :

A closed circular tube of average radius 15 cm, whose inner walls are rough, is kept in a vertical plane. A block of mass 1 kg just fits inside the tube. The speed of the block is 22 m/s when it is introduced at the top of the tube. After completing five oscillations, the block stops at the bottom region of the tube. The work done by the tube on the block is ______ J. [Given $g = 10$ $m / s^{2}$ ] [2023]

[IMAGE 40]

0%

(245)

$r_{avg} = 15 cm$

$W_{f} + W_{g} = Δ K E$

$W_{f} + 10 \times 0.3 = - \frac{1}{2} \times 484$

$W_{f} = - 245 J$

Q 18 :

To maintain a speed of 80 km/h by a bus of mass 500 kg on a plane rough road for 4 km distance, the work done by the engine of the bus will be ______ kJ. (The coefficient of friction between tyre of bus and road is 0.04.) [2023]

0%

(784)

$For constant speed, W D by engine + W D by friction = 0 [by WET]$

$W D_{engine} = - W D_{friction} = - [- μ m g x]$

$= 0.04 \times 500 \times 9.8 \times 4 \times 10^{3} = 784 kJ$

Q 19 :

A car accelerates from rest to $u$ m/s. The energy spent in this process is E J. The energy required to accelerate the car from $u$ m/s to $2 u$ m/s is $n E$ J. The value of $n$ is ______. [2023]

0%

(3)

$E_{1} = \frac{1}{2} m u^{2} - 0 = \frac{1}{2} m u^{2} = E$

$E_{2} = \frac{1}{2} m {(2 u)}^{2} - \frac{1}{2} m u^{2} = \frac{3}{2} m u^{2} = 3 E$

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