Work Power & Energy Questions — JEE Physics

Q1

A particle of mass

100g

is thrown vertically upwards with a speed of

5

m/s

. The work done by the force of gravity during the time the particle goes up is

A

-0.5J

B

-1.25J

C

1.25J

D

0.5J

Correct Answer

Option B

Solution

Kinetic energy at point of throwing is converted into potential energy of the particle during rise.

K.E = {1 \over 2}m{v^2} = {1 \over 2} \times 0.1 \times 25 = 1.25\,J

W = - mgh = - \left( {{1 \over 2}m{v^2}} \right) = - 1.25\,J

\left[ \, \right.

As we know,

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

by energy conservation

\left. \, \right]

Q2

A bullet of mass

0.1 \mathrm{~kg}

moving horizontally with speed

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

hits a wooden block of mass

3.9 \mathrm{~kg}

kept on a horizontal rough surface. The bullet gets embedded into the block and moves

20 \mathrm{~m}

before coming to rest. The coefficient of friction between the block and the surface is __________. (Given

g=10 \mathrm{~m} / \mathrm{s}^{2}

)

A 0.65

B 0.25

C 0.50

D 0.90

Correct Answer

Option B

Solution

First, we will use conservation of momentum to find the velocity of the bullet-block system just after the bullet gets embedded into the block.

The initial momentum of the system is given by the momentum of the bullet (as the block is initially at rest), and the final momentum of the system is the combined momentum of the bullet and the block.

Setting initial momentum equal to final momentum: $m_{\text{bullet}} \cdot v_{\text{bullet}} = (m_{\text{bullet}} + m_{\text{block}}) \cdot v_{\text{final}}$ Solving for ( $v_{\text{final}}$ ): $v_{\text{final}} = \dfrac{m_{\text{bullet}} \cdot v_{\text{bullet}}}{m_{\text{bullet}} + m_{\text{block}}}$ Substituting the given values: $v_{\text{final}} = \dfrac{0.1 \, \text{kg} \cdot 400 \, \text{m/s}}{0.1 \, \text{kg} + 3.9 \, \text{kg}} = 10 \, \text{m/s}$ Next, we know the block comes to rest after moving 20 m due to friction.

The work done by the friction force is equal to the initial kinetic energy of the block (since it comes to rest, the final kinetic energy is 0).

The work done by friction is given by the friction force times the distance, and the friction force is equal to the coefficient of friction times the normal force (which is equal to the weight of the block).

So, setting the work done by friction equal to the initial kinetic energy of the block: $\mu \cdot (m_{\text{bullet}} + m_{\text{block}}) \cdot g \cdot d = \dfrac{1}{2} \cdot (m_{\text{bullet}} + m_{\text{block}}) \cdot v_{\text{final}}^2$ Solving for ( $\mu$ ): $\mu = \dfrac{\dfrac{1}{2} \cdot (m_{\text{bullet}} + m_{\text{block}}) \cdot v_{\text{final}}^2}{(m_{\text{bullet}} + m_{\text{block}}) \cdot g \cdot d}$ Substituting the given values: $\mu = \dfrac{\dfrac{1}{2} \cdot (0.1 \, \text{kg} + 3.9 \, \text{kg}) \cdot (10 \, \text{m/s})^2}{(0.1 \, \text{kg} + 3.9 \, \text{kg}) \cdot 10 \, \text{m/s}^2 \cdot 20 \, \text{m}} = 0.25$

Q3

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particles takes place in a plane. It follows that

A its kinetic energy is constant

B is acceleration is constant

C its velocity is constant

D it moves in a straight line

Correct Answer

Option A

Solution

Work done by such force is always zero when a force of constant magnitude always at right angle to the velocity of a particle when the motion of the particle takes place in a plane.

$\therefore$ From work-energy theorem,

\Delta K = 0

$\therefore$

K

remains constant.

Q4

This question has Statement

1

and Statement

2.

Of the four choices given after the Statements, choose the one that best describes the two Statements. If two springs

{S_1}

and

{S_2}

of force constants

{k_1}

and

{k_2}

, respectively, are stretched by the same force, it is found that more work is done on spring

{S_1}

than on spring

{S_2}

. STATEMENT 1: If stretched by the same amount work done on

{S_1}

, Work done on

{S_1}

is more than

{S_2}

STATEMENT 2:

{k_1} < {k_2}

A Statement 1 is false, Statement 2 is true

B Statement 1 is true, Statement 2 is false

C Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1

D Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1

Correct Answer

Option A

Solution

We know force (F) = kx

W = {1 \over 2}k{x^2}

W =

{{{{\left( {kx} \right)}^2}} \over {2k}}

\,\,\,

$\therefore$

W = {{{F^2}} \over {2k}}

[ as

F=kx

] When force is same then,

W \propto {1 \over k}

Given that,

{W_1} > {W_2}

$\therefore$

{k_1} < {k_2}

Statement-2 is true. For the same extension, x1 = x2 = x Work done on spring S1 is W1 =

{1 \over 2}{k_1}x_1^2 = {1 \over 2}{k_1}{x^2}

Work done on spring S2 is W2 =

{1 \over 2}{k_2}x_2^2 = {1 \over 2}{k_2}{x^2}

$\therefore$

{{{W_1}} \over {{W_2}}} = {{{k_1}} \over {{k_2}}}

As

{k_1} < {k_2}

then

{W_1} < {W_2}

So, Statement-1 is false.

Q5

A spring of spring constant

5 \times {10^3}\,N/m

is stretched initially by

5

cm

from the unstretched position. Then the work required to stretch it further by another

5

cm

is

A

12.50

N

-

m

B

18.75

N

-

m

C

25.00

N

-

m

D

625

N

-

m

Correct Answer

Option B

Solution

Given

k = 5 \times {10^3}N/m

Work done when a spring stretched from x1 cm to x2 cm,

W = {1 \over 2}k\left( {x_2^2 - x_1^2} \right)

= {1 \over 2} \times 5 \times {10^3}\left[ {{{\left( {0.1} \right)}^2} - {{\left( {0.05} \right)}^2}} \right]

= {{5000} \over 2} \times 0.15 \times 0.05 = 18.75\,\,Nm

Q6

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

45^\circ

to the horizontal. The kinetic energy of the ball at the highest point of its height will be

A E

B

{E \over {\sqrt 2 }}

C

{E \over 2}

D zero

Correct Answer

Option C

Solution

Assume the ball of mass m is projected with a speed u. Then the kinetic energy(E) at the point of projection =

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

At highest point of flight only horizontal component of velocity

u\cos \theta

present as at highest point vertical component of velocity is = 0.

Note : The horizontal component of velocity does not change in entire projectile motion.

At highest point the velocity is =

u\cos \theta

=

u\cos 45^\circ

=

{u \over {\sqrt 2 }}

$\therefore$ The kinetic energy at the height point =

{1 \over 2}m{\left( {{u \over {\sqrt 2 }}} \right)^2}

=

{1 \over 2}m{u^2} \times {1 \over 2}

=

{E \over 2}

Q7

A uniform chain of length

2

m

is kept on a table such that a length of

60

cm

hangs freely from the edge of the table. The total mass of the chain is

4

kg.

What is the work done in pulling the entire chain on the table?

A

12

J

B

3.6

J

C

7.2

J

D

1200

J

Correct Answer

Option B

Solution

Mass of hanging part

(m') = {4 \over 2} \times \left( {0.6} \right)kg

= 1.2 kg Let at the surface

PE=0

Center of mass of hanging part

=0.3

m

below the surface of the table

{U_i} = - m'gx = - 1.2 \times 10 \times 0.30

= - 3.6 J

\Delta U = m'gx = 3.6 J =

Work done in putting the entire chain on the table.

Q8

A force

\overrightarrow F = \left( {5\overrightarrow i + 3\overrightarrow j + 2\overrightarrow k } \right)N

is applied over a particle which displaces it from its origin to the point

\overrightarrow r = \left( {2\overrightarrow i - \overrightarrow j } \right)m.

The work done on the particle in joules is

A

+10

B

+7

C

-7

D

+13

Correct Answer

Option B

Solution

The work done by a force on a particle is given by the dot product of the force and the displacement vector of the particle:

W = \overrightarrow F \cdot \overrightarrow r

We can substitute the given vectors into this expression:

W = \overrightarrow F .\overrightarrow r

= \left( {5\widehat i + 3\widehat j + 2\widehat k} \right).\left( {2\widehat i - \widehat j} \right)

=10-3=7

J

Q9

A body of mass

' m ',

acceleration uniformly from rest to

'{v_1}'

in time

{T}

. The instantaneous power delivered to the body as a function of time is given by

A

{{m{v_1}{t^2}} \over {{T}}}

B

{{mv_1^2t} \over {T^2}}

C

{{m{v_1}t} \over {{T}}}

D

{{mv_1^2t} \over {{T}}}

Correct Answer

Option B

Solution

Assume acceleration of body be

a

$\therefore$

{v_1} = 0 + a{T} \Rightarrow a = {{{v_1}} \over {{T}}}

$\therefore$

v = at \Rightarrow v = {{{v_1}t} \over {{T}}}

{P_{inst}} = \overrightarrow F .\overrightarrow v = \left( {m\overrightarrow a } \right).\overrightarrow v

= \left( {{{m{v_1}} \over {{T}}}} \right)\left( {{{{v_1}t} \over {{T}}}} \right)

= m{\left( {{{{v_1}} \over {{T}}}} \right)^2}t

Q10

A bullet fired into a fixed target loses half of its velocity after penetrating

3

cm.

How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?

A

2.0

cm

B

3.0

cm

C

1.0

cm

D

1.5

cm

Correct Answer

Option C

Solution

Let

K

be the initial kinetic energy and

F

be the resistive force. Then according to work-energy theorem,

W = \Delta K

$ i.e.,

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

Let the bullet will penetrate x cm more before coming to rest. $\therefore$

Fx = {1 \over 2}m{\left( {{v \over 2}} \right)^2} - {1 \over 2}m{\left( 0 \right)^2}...\left( 2 \right)

Dividing eq.

(1)

and

(2)

we get,

{x \over 3} = {1 \over 3}

or x = 1 cm