Work Power & Energy — Page 3 | JEE Physics

Q21

A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time 't' is proportional to :-

A t2/3

B t3/2

C t

D t1/2

Correct Answer

Option B

Solution

P = F.v = mav

P = {{mvdv} \over {dt}}

\int\limits_0^t {Pdt} = m\int\limits_0^v {vdv}

Pt = {{m{v^2}} \over 2}

v = \sqrt {{{2Pt} \over m}}

{{dx} \over {dt}} = \sqrt {{{2Pt} \over m}}

\int {dx} = \int {\sqrt {{{2Pt} \over m}} } dt

x \propto {t^{3/2}}

Q22

A body at rest is moved along a horizontal straight line by a machine delivering a constant power. The distance moved by the body in time 't' is proportional to :

A

{t^{{3 \over 2}}}

B

{t^{{1 \over 2}}}

C

{t^{{1 \over 4}}}

D

{t^{{3 \over 4}}}

Correct Answer

Option A

Solution

P = constant

{1 \over 2}

mv2 = Pt $\Rightarrow$ v $\propto$

\sqrt t

{{dx} \over {dt}} = C\sqrt t

[C = constant] by integration.

x = C{{{t^{{1 \over 2} + 1}}} \over {{1 \over 2} + 1}}

x \propto {t^{3/2}}

Q23

A body of mass 'm' dropped from a height 'h' reaches the ground with a speed of 0.8

\sqrt {gh}

. The value of workdone by the air-friction is :

A

-

0.68 mgh

B mgh

C 1.64 mgh

D 0.64 mgh

Correct Answer

Option A

Solution

Given, the mass of the body = m The height from which the body dropped = h The speed of the body when reached the ground,

{v_f} = 0.8\sqrt {gh}

Initial velocity of the body, v = 0 m/s Using the work-energy theorem, Work done by gravity + Work done by air-friction = Final kinetic energy $-$ Initial kinetic energy.

{W_{mg}} + {W_{air - friction}} = {1 \over 2}mv_f^2 - {1 \over 2}mv_i^2

Here, work done by gravity = mgh

\Rightarrow mgh + {W_{air - friction}} = {1 \over 2}m{(0.8\sqrt {gh} )^2} - {1 \over 2}m{(0)^2}

\Rightarrow {W_{air - friction}} = {{0.64mgh} \over 2} - mgh

\Rightarrow 0.32mgh - mgh = - 0.68mgh

The value of the work done by the air friction is $-$ 0.68 mgh.

Q24

A particle of mass 500 gm is moving in a straight line with velocity v = b x5/2. The work done by the net force during its displacement from x = 0 to x = 4 m is : (Take b = 0.25 m

-

3/2 s

-

1).

A 2 J

B 4 J

C 8 J

D 16 J

Correct Answer

Option D

Solution

{W_{total}} = \Delta K

= {1 \over 2}\left( {{1 \over 2}} \right)\left[ {{{\{ b{{(4)}^{5/2}}\} }^2} - 0} \right]

= {{{b^2}} \over 4} \times {4^5}

\Rightarrow {W_{total}} = 16\,J

Q25

A body of mass

0.5 \mathrm{~kg}

travels on straight line path with velocity

v=\left(3 x^{2}+4\right) \mathrm{m} / \mathrm{s}

. The net workdone by the force during its displacement from

x=0

to

x=2 \mathrm{~m}

is :

A 64 J

B 60 J

C 120 J

D 128 J

Correct Answer

Option B

Solution

v = 3{x^2} + 4

at

x = 0

,

{v_1} = 4

m/s

x = 2

,

{v_2} = 16

m/s $\Rightarrow$ Work done =

\Delta

kinetic energy

= {1 \over 2} \times m\left( {v_2^2 - v_1^2} \right)

= {1 \over 4}(256 - 16)

= 60

J

Q26

A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of 200 g travelling with speed 10 ms

-

1 gets embedded in it, then loss of kinetic energy will be :

A 4.9 J

B 9.8 J

C 14.7 J

D 19.6 J

Correct Answer

Option B

Solution

Loss in

KE = {1 \over 2} \times {{{m_1}{m_2}} \over {{m_1} + {m_2}}} \times {v^2}

= {1 \over 2} \times {{9.8 \times 0.2} \over {10}} \times {(10)^2}

= 9.8

J

Q27

Sand is being dropped from a stationary dropper at a rate of

0.5 \,\mathrm{kgs}^{-1}

on a conveyor belt moving with a velocity of

5 \mathrm{~ms}^{-1}

. The power needed to keep the belt moving with the same velocity will be :

A 1.25 W

B 2.5 W

C 6.25 W

D 12.5 W

Correct Answer

Option D

Solution

{{dm} \over {dt}} = 0.5

kg/s

v = 5

m/s

F = {{vdm} \over {dt}} = 2.5

kg m/s2

P = \overline F \,.\,\overline v = (2.5)(5)

W

= 12.5

W

Q28

A ball is projected with kinetic energy E, at an angle of

60^{\circ}

to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :

A Zero

B

\dfrac{E}{2}

C

\dfrac{E}{4}

D E

Correct Answer

Option C

Solution

K.E. = E = {1 \over 2}m{v^2}

at highest point

K.E' = {1 \over 2}m{v^2}{\cos ^2}\theta

= {1 \over 2}m{v^2}\left( {{1 \over 4}} \right)

= {E \over 4}

Q29

A stone is projected at angle

30^{\circ}

to the horizontal. The ratio of kinetic energy of the stone at point of projection to its kinetic energy at the highest point of flight will be -

A 1 : 4

B 1 : 2

C 4 : 3

D 4 : 1

Correct Answer

Option C

Solution

\mathrm{KE}_{\mathrm{in}}=\frac{1}{2} m v^{2}

$\mathrm{KE}_{\text{final }}=\dfrac{1}{2} m v^{2} \cos ^{2} 30^{\circ}=\dfrac{1}{2} m v^{2}\left(\dfrac{\sqrt{3}}{2}\right)^{2}$ $\dfrac{\mathrm{KE}_{\mathrm{in}}}{\mathrm{KE}_{\mathrm{f}}}=\dfrac{\dfrac{1}{2} m v^{2}}{\dfrac{1}{2} m v^{2}\left(\dfrac{3}{4}\right)}=\dfrac{4}{3}$

Q30

The ratio of powers of two motors is

\dfrac{3 \sqrt{x}}{\sqrt{x}+1}

, that are capable of raising

300 \mathrm{~kg}

water in 5 minutes and

50 \mathrm{~kg}

water in 2 minutes respectively from a well of

100 \mathrm{~m}

deep. The value of

x

will be

A 16

B 4

C 2

D 2.4

Correct Answer

Option A

Solution

Let us first find the power required to lift the water using each motor.

Let $P_1$ be the power of the first motor, and $P_2$ be the power of the second motor.

The work done in lifting the water is given by $W = mgh$ , where $m$ is the mass of water lifted, $g$ is the acceleration due to gravity, and $h$ is the height through which the water is lifted.

In this case, $m = 300\mathrm{~kg}$ and $h = 100\mathrm{~m}$ for the first motor, and $m = 50\mathrm{~kg}$ and $h = 100\mathrm{~m}$ for the second motor.

The work done in lifting the water in 5 minutes by the first motor is:

W_1 = mgh = (300\mathrm{~kg})(9.8\mathrm{~m/s^2})(100\mathrm{~m}) = 294000\mathrm{~J}

The power required to do this work in 5 minutes is:

P_1 = \frac{W_1}{t_1} = \frac{294000\mathrm{~J}}{300\mathrm{~s}} = 980\mathrm{~W}

The work done in lifting the water in 2 minutes by the second motor is:

W_2 = mgh = (50\mathrm{~kg})(9.8\mathrm{~m/s^2})(100\mathrm{~m}) = 49000\mathrm{~J}

The power required to do this work in 2 minutes is:

P_2 = \frac{W_2}{t_2} = \frac{49000\mathrm{~J}}{120\mathrm{~s}} = 408.33\mathrm{~W}

The ratio of the powers of the two motors is:

\frac{P_1}{P_2} = \frac{980\mathrm{~W}}{408.33\mathrm{~W}} \approx 2.4

We are given that this ratio is equal to:

\frac{3 \sqrt{x}}{\sqrt{x}+1}

We can solve for $x$ as follows:

\frac{3 \sqrt{x}}{\sqrt{x}+1} = 2.4

3\sqrt{x} = 2.4(\sqrt{x}+1)

3\sqrt{x} = 2.4\sqrt{x} + 2.4

(3-2.4)\sqrt{x} = 2.4

0.6\sqrt{x} = 2.4

\sqrt{x} = 4

x = 16

Therefore, the value of $x$ is 16.