Work Power & Energy — Page 6 | JEE Physics

Q51

A particle which is experiencing a force, given by

\overrightarrow F = 3\widehat i - 12\widehat j,

undergoes a displacement of

\overrightarrow d = 4\overrightarrow i

particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement ?

A 9 J

B 10 J

C 12 J

D 15 J

Correct Answer

Option D

Solution

Work done =

\overrightarrow F \cdot \overrightarrow d

$=$ 12 J work energy theorem wnet $=$

\Delta

K.E. 12 $=$ Kf $-$ 3 Kf = 15 J

Q52

A body is moving unidirectionally under the influence of a constant power source. Its displacement in time t is proportional to :

A t2/3

B t3/2

C t

D t2

Correct Answer

Option B

Solution

When a body moves under the influence of a constant power, the relationship between displacement and time can be established through the concept of power.

Power (P) is defined as the rate at which work is done, and it can also be expressed in terms of force (F) and velocity (v) as $P = F \cdot v$ .

For a constant power P and assuming the force acts in the direction of the velocity, we can analyze how displacement (s) changes with time (t).

Since force can also be written as $F = \dfrac{d(mv)}{dt}$ for a constant mass m, this simplifies to $F = m \dfrac{dv}{dt}$ , because mass doesn't change with time for most cases.

Integrating force over a distance gives work (W), and power is the rate of doing work, thus we can connect these concepts.

The kinetic energy (K.E) of the body is given by $K.E = \dfrac{1}{2}mv^2$ , and the work done by the force is equal to the change in kinetic energy.

Considering power is constant, $P = \dfrac{dW}{dt} = \dfrac{d(\dfrac{1}{2}mv^2)}{dt}$ .

Rearranging terms to focus on velocity and integrating with respect to time will give us a relation involving velocity and time.

For a constant mass system, and using $P = F \cdot v = m \cdot a \cdot v = m \cdot \dfrac{dv}{dt} \cdot v$ , and knowing that $P = \text{constant}$ , we rearrange to find the relationship between velocity and time.

Given $P = m \cdot v \cdot \dfrac{dv}{dt}$ , we rearrange to $\dfrac{P}{m} dt = v dv$ .

Integrating both sides where the initial condition is when $t = 0, v = 0$ , we get $\dfrac{P}{m} t = \dfrac{1}{2} v^2$ , solving for $v$ gives $v \propto t^{1/2}$ , so $v = k \cdot t^{1/2}$ for some constant $k$ .

The displacement $s$ is obtained by integrating the velocity with respect to time, $s = \int v dt = \int k \cdot t^{1/2} dt = \dfrac{2}{3}k \cdot t^{3/2}$ .

Therefore, the displacement $s$ is proportional to $t^{3/2}$ .

The correct answer is Option B, $t^{3/2}$ .

Q53

A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time

't'

is proportional to

A

{t^{3/4}}

B

{t^{3/2}}

C

{t^{1/4}}

D

{t^{1/2}}

Correct Answer

Option B

Solution

We know that

F \times v =

Power According to the question, power is constant. $\therefore$

F \times v = c\,\,\,\,

where

c=

constant $\therefore$

m{{dv} \over {dt}} \times v = c

\,\,\,\,\left( \, \right.

$\therefore$

\left. {F = ma = {{mdv} \over {dt}}\,\,} \right)

$\therefore$

m\int\limits_0^v {vdv = c\int\limits_0^t {dt} } \,\,\,\,\,\,\,\,\,

$\therefore$

{1 \over 2}m{v^2} = ct

$\therefore$

v = \sqrt {{{2c} \over m}} \times {t^{{{1}/{2}}}}

{{dx} \over {dt}} = \sqrt {{{2c} \over m}} \times {t^{{{1}/{2}}}}\,\,\,\,

where

v = {{dx} \over {dt}}

$\therefore$

\int\limits_0^x {dx = \sqrt {{{2c} \over m}} } \times \int\limits_0^t {{t^{{{1}/{2}}}}} dt

x = \sqrt {{{2c} \over m}} \times {{2{t^{{{3}/{2}}}}} \over 3} \Rightarrow x \propto {t^{{{3}/{2}}}}

Q54

A person trying to lose weight by burning fat lifts a mass of

10

kg

upto a height of

1

m

1000

times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies

3.8 \times {10^7}J

of energy per

kg

which is converted to mechanical energy with a

20\%

efficiency rate. Take

g = 9.8\,m{s^{ - 2}}

:

A

9.89 \times {10^{ - 3}}\,\,kg

B

12.89 \times {10^{ - 3}}\,kg

C

2.45 \times {10^{ - 3}}\,\,kg

D

6.45 \times {10^{ - 3}}\,\,kg

Correct Answer

Option B

Solution

Assume the amount of fat is used = x kg So total Mechanical energy available through fat =

x \times 3.8 \times {10^7} \times {{20} \over {100}}

And work done through lifting up = 10 $\times$ 9.8 $\times$ 1000 = 98000 J $\Rightarrow$

x \times 3.8 \times {10^7} \times {{20} \over {100}}

= 98000 $\Rightarrow$

x

= 12.89 $\times$ 10-3 kg

Q55

Which one of the following forces cannot be expressed in terms of potential energy?

A Frictional force

B Coulomb’s force

C Restoring force

D Gravitational force

Correct Answer

Option A

Solution

The correct answer is:

\textbf{Option A: Frictional force}

Here's why: Conservative vs.

Non-Conservative Forces: Conservative Forces: These are forces for which the work done is independent of the path taken; they only depend on the initial and final positions.

For conservative forces, you can define a potential energy function.

Examples include Coulomb’s force, gravitational force, and the restoring force of a spring.

Non-Conservative Forces: These forces depend on the path taken and usually result in energy dissipation (often as heat).

Friction is a typical example, and hence, it cannot be described solely by a potential energy function.

Example Explanations: Coulomb’s Force: It is a conservative force because the work done in moving a charge in an electric field only depends on the difference in electric potential between two points.

Gravitational Force: It is also conservative.

The potential energy associated with gravitational force is given by gravitational potential energy, for example,

U = mgh

near the Earth’s surface. Restoring Force: For a spring, the force is given by Hooke’s law:

F = -kx

and the corresponding potential energy is

U = \frac{1}{2}kx^2.

This is a conservative force.

Since friction is a non-conservative force that dissipates energy, it cannot be expressed in terms of potential energy.

Therefore, the correct answer is Option A: Frictional force.

Q56

An automobile of mass 'm' accelerates starting from origin and initially at rest, while the engine supplies constant power P. The position is given as a function of time by :

A

{\left( {{{9P} \over {8m}}} \right)^{{1 \over 2}}}{t^{{3 \over 2}}}

B

{\left( {{{8P} \over {9m}}} \right)^{{1 \over 2}}}{t^{{2 \over 3}}}

C

{\left( {{{9m} \over {8P}}} \right)^{{1 \over 2}}}{t^{{3 \over 2}}}

D

{\left( {{{8P} \over {9m}}} \right)^{{1 \over 2}}}{t^{{3 \over 2}}}

Correct Answer

Option D

Solution

P = const.

P = Fv = {{m{v^2}dv} \over {dx}}

\int\limits_0^x {{P \over m}dx} = \int\limits_0^v {{v^2}dv}

{{Px} \over m} = {{{v^3}} \over 3}

{\left( {{{3Px} \over m}} \right)^{1/3}} = v = {{dx} \over {dt}}

{\left( {{{3P} \over m}} \right)^{1/3}}\int\limits_0^t {dt} = \int\limits_0^x {{x^{ - 1/3}}} dx

\Rightarrow x = {\left( {{{8P} \over {9m}}} \right)^{1/2}}{t^{3/2}}

Q57

If a body looses half of its velocity on penetrating

3

cm

in a wooden block, then how much will it penetrate more before coming to rest?

A

1

cm

B

2

cm

C

3

cm

D

4

cm

Correct Answer

Option A

Solution

We know the work energy theorem,

W = \Delta K = FS

For first penetration, by applying work energy theorem we get,

{1 \over 2}m{v^2} - {1 \over 2}m{\left( {{v \over 2}} \right)^2} = F \times 3\,\,...(i)

For second penetration, by applying work energy theorem we get,

{1 \over 2}m{\left( {{v \over 2}} \right)^2} - 0 = F \times S\,...(ii)

On dividing

(ii)

by

(i)

{{1/4} \over {3/4}} = S/3

$\therefore$

S = 1\,cm

Q58

A particle is moving in a circular path of radius

a

under the action of an attractive potential

U = - {k \over {2{r^2}}}

Its total energy is:

A

- {3 \over 2}{k \over {{a^2}}}

B Zero

C

- {k \over {4{a^2}}}

D

{k \over {2{a^2}}}

Correct Answer

Option B

Solution

We know, Total energy = Kinetic energy + Potential energy Potential energy given as

U = - {k \over {2{r^2}}}

We need to find Kinetic Energy. As Force acting on the particle (F) =

- {{dU} \over {dr}}

\Rightarrow F = - {d \over {dr}}\left( {{{ - k} \over {2{r^2}}}} \right)

= {k \over 2} \times \left( { - 2} \right) \times {r^{ - 3}}

= - {k \over {{r^3}}}

Because of this force particle is having circular motion so it will provide possible centripetal force.

\left| F \right| = {{m{v^2}} \over r}

\Rightarrow {{m{v^2}} \over r} = {k \over {{r^3}}}

$\Rightarrow$

m{v^2} = {k \over {{r^2}}}

We know kinetic energy of particle, K =

{1 \over 2}m{v^2}

=

{k \over {2{r^2}}}

As Total energy = Kinetic energy + Potential energy So Total energy =

{k \over {2{r^2}}}

- {k \over {2{r^2}}}

= 0

Q59

A boy is rolling a 0.5 kg ball on the frictionless floor with the speed of 20 ms-1. The ball gets deflected by an obstacle on the way. After deflection it moves with 5% of its initial kinetic energy. What is the speed of the ball now?

A 14.41 ms

-

1

B 19.0 ms

-

1

C 4.47 ms

-

1

D 1.00 ms

-

1

Correct Answer

Option C

Solution

K.E{._f} = 5\% \,K{E_i}

{1 \over 2}m{v^2} = {5 \over {100}} \times {1 \over 2} \times m \times {20^2}

{v^2} = {1 \over {20}} \times {20^2} = 20

v = \sqrt {20} = 2\sqrt 5

m/s = 4.47 m/s

Q60

A spherical ball of mass

20

kg

is stationary at the top of a hill of height

100

m

. It rolls down a smooth surface to the ground, then climbs up another hill of height

30

m

and finally rolls down to a horizontal base at a height of

20

m

above the ground. The velocity attained by the ball is

A

20

m/s

B

40

m/s

C

10\sqrt {30} \,\,\,m/s

D

10\,\,m/s

Correct Answer

Option B

Solution

Loss in potential energy $=$ gain in kinetic energy

m \times g \times 80 = {1 \over 2}m{v^2}

$\Rightarrow$

10 \times 80 = {1 \over 2}{v^2}

$\Rightarrow$

{v^2} = 1600

or

v = 40\,m/s