Work Power & Energy — Page 5 | JEE Physics

Q41

A block of mass 25 kg is pulled along a horizontal surface by a force at an angle

45^{\circ}

with the horizontal. The friction coefficient between the block and the surface is 0.25 . The block travels at a uniform velocity. The work done by the applied force during a displacement of 5 m of the block is :

A 490 J

B 970 J

C 735 J

D 245 J

Correct Answer

Option D

Solution

Block travels with uniform velocity So $\quad \mathrm{a}=0 \Rightarrow \mathrm{~F} \cos 45^{\circ}=$ friction

\begin{aligned} & \frac{\mathrm{F}}{\sqrt{2}}=\mu\left[\mathrm{mg}-\frac{\mathrm{F}}{\sqrt{2}}\right] \\ & \frac{\mathrm{F}}{\sqrt{2}}=0.25\left[25 \times 9.8-\frac{\mathrm{F}}{\sqrt{2}}\right] \\ & \Rightarrow \quad 1.25 \frac{\mathrm{~F}}{\sqrt{2}}=61.25 \\ & \mathrm{~F}=\frac{61.25 \times \sqrt{2}}{1.25}=49 \sqrt{2} \\ & \mathrm{~W}_{\mathrm{ext}}=\mathrm{FS} \cos 45^{\circ} \\ &=49 \sqrt{2} \times 5 \times \frac{1}{\sqrt{2}}=245 \mathrm{~J} \end{aligned}

Q42

A car of weight W is on an inclined road that rises by 100 m over a distance of 1 km and applies a constant frictional force

{W \over 20}

on the car. While moving uphill on the road at a speed of 10 ms−1, the car needs power P. If it needs power

{p \over 2}

while moving downhill at speed v then value of

\upsilon

is :

A 20 ms

-

1

B 15 ms

-

1

C 10 ms

-

1

D 5 ms

-

1

Correct Answer

Option B

Solution

Here, tan $\theta$ =

{{100} \over {1000}} = {1 \over {10}}

$\therefore$ sin $\theta$ =

{1 \over {10}}

(as $\theta$ is very small), when car is moving uphill : P = f $\times$ u = (wsin $\theta$ + f) $\times$ u =

\left( {{w \over {10}} + {w \over {20}}} \right) \times 10

P =

{{3w} \over {20}} \times 10

=

{{3w} \over 2}

When car is moving down hill : $\therefore$

{P \over 2}

= (wsin $\theta$ $-$ f) $\times$ v $\Rightarrow$

{{3w} \over 4}

=

\left( {{w \over {10}} - {w \over {20}}} \right)

$\times$ v $\Rightarrow$

{{w \over {20}} \times }

v =

{{3w} \over 4}

$\Rightarrow$ v = 15 m/s

Q43

A force acts on a 2 kg object so that its position is given as a function of time as x = 3t2 + 5. What is the work done by this force in first 5 seconds ?

A 850 J

B 950 J

C 875 J

D 900 J

Correct Answer

Option D

Solution

Displacement, x = 3t2 + 5 $\therefore$ v =

{{dx} \over {dt}} = 6t

At t = 0, velocity = 6 $\times$ 0 = 0 at t = 5, velocity = 5 $\times$ 6 = 30 m/s we know from work energy theorem, Work (W) = change in kinetic energy (

\Delta

K) =

{1 \over 2}mv_F^2 - {1 \over 2}mv_i^2

=

{1 \over 2}

$\times$ 2 $\times$ (30)2 $-$ 0 = 900 J

Q44

The potential energy function for the force between two atoms in a diatomic molecule is approximately given by

U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}},

where

a

and

b

are constants and

x

is the distance between the atoms. If the dissociation energy of the molecule is

D = \left[ {U\left( {x = \infty } \right) - {U_{at\,\,equilibrium}}} \right],\,\,D

is

A

{{{b^2}} \over {2a}}

B

{{{b^2}} \over {12a}}

C

{{{b^2}} \over {4a}}

D

{{{b^2}} \over {6a}}

Correct Answer

Option C

Solution

Given

U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}}

{U\left( {x = \infty } \right)}

= 0 We know

F = - {{dU} \over {dx}} = - \left[ {{{12a} \over {{x^{13}}}} + {{6b} \over {{x^7}}}} \right]

At equilibrium:

{{dU\left( x \right)} \over {dx}} = 0

\Rightarrow {{ - 12a} \over {{x^{13}}}} = {{ - 6b} \over {{x^7}}}

\Rightarrow x = {\left( {{{2a} \over h}} \right)^{{1 \over 6}}}

$\therefore$

{U_{at\,\,equilibrium\,}} = {a \over {{{\left( {{{2a} \over b}} \right)}^2}}} - {b \over {\left( {{{2a} \over b}} \right)}}

= - {{{b^2}} \over {4a}}

$\therefore$

D = 0 - \left( { - {{{b^2}} \over {4a}}} \right) = {{{b^2}} \over {4a}}

Q45

When a rubber-band is stretched by a distance

x

, it exerts restoring force of magnitude

F = ax + b{x^2}

where

a

and

b

are constants. The work done in stretching the unstretched rubber-band by

L

is :

A

a{L^2} + b{L^3}

B

{1 \over 2}\left( {a{L^2} + b{L^3}} \right)

C

{{a{L^2}} \over 2} + {{b{L^3}} \over 3}

D

{1 \over 2}\left( {{{a{L^2}} \over 2} + {{b{L^3}} \over 3}} \right)

Correct Answer

Option C

Solution

Given Restoring force, F = ax + bx2 Work done in stretching the rubber-band by a distance

dx

is

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dW = F\,dx = \left( {ax + b{x^2}} \right)dx

Intergrating both sides,

W = \int\limits_0^L {axdx + \int\limits_0^L {b{x^2}dx}}

=

\left[ {a{{{x^2}} \over 2} + b{{{x^3}} \over 3}} \right]_0^L

=

{{a{L^2}} \over 2} + {{b{L^3}} \over 3}

Q46

A bullet is fired into a fixed target looses one third of its velocity after travelling

4 \mathrm{~cm}

. It penetrates further

\mathrm{D} \times 10^{-3} \mathrm{~m}

before coming to rest. The value of

\mathrm{D}

is :

A 23

B 32

C 42

D 52

Correct Answer

Option B

Solution

\begin{aligned} & v^2-u^2=2 a S \\ & \left(\frac{2 u}{3}\right)^2=u^2+2(-a)\left(4 \times 10^{-2}\right) \\ & \frac{4 u^2}{9}=u^2-2 a\left(4 \times 10^{-2}\right) \\ & -\frac{5 u^2}{9}=-2 a\left(4 \times 10^{-2}\right) \ldots(1) \\ & 0=\left(\frac{2 u}{3}\right)^2+2(-a)(x) \\ & -\frac{4 u^2}{9}=-2 a x \ldots(2) \end{aligned}

(1)/(2)

\begin{aligned} & \frac{5}{4}=\frac{4 \times 10^{-2}}{\mathrm{x}} \\ & \mathrm{x}=\frac{16}{5} \times 10^{-2} \\ & \mathrm{x}=3 \cdot 2 \times 10^{-2} \mathrm{~m} \\ & \mathrm{x}=32 \times 10^{-3} \mathrm{~m} \end{aligned}

Q47

A particle is released from height

S

above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.

A

\dfrac{\mathrm{S}}{4}, \dfrac{3 \mathrm{gS}}{2}

B

\dfrac{\mathrm{S}}{2}, \dfrac{3 \mathrm{gS}}{2}

C

\dfrac{\mathrm{S}}{4}, \sqrt{\dfrac{3 \mathrm{gS}}{2}}

D

\dfrac{\mathrm{S}}{2}, \sqrt{\dfrac{3 \mathrm{gS}}{2}}

Correct Answer

Option C

Solution

Using the equation of motion for velocity at height $x$ : $V^2 = 0 + 2g(S - x)$ Simplifies to: $V^2 = 2g(S - x)$ Potential Energy at height $x$ : $\text{Potential Energy} = mgx$ Relation between kinetic and potential energy: Given that the kinetic energy is three times the potential energy, $mgx = 3 \times \dfrac{1}{2} mV^2$ Simplifying gives: $gx = \dfrac{3}{2} \times 2g(S - x)$ $gx = 3g(S - x)$ $4x = S$ Solving for $x$ : $x = \dfrac{S}{4}$ Finding the speed $V$ at this instant: $V = \sqrt{2g \left( S - \dfrac{S}{4} \right)} = \sqrt{2g \times \dfrac{3S}{4}} = \sqrt{\dfrac{3gS}{2}}$ Thus, the height from the surface of the Earth is $\dfrac{S}{4}$ and the speed of the particle at that height is $\sqrt{\dfrac{3gS}{2}}$ .

Q48

A spring of force constant

800

N/m

has an extension of

5

cm.

The work done in extending it from

5

cm

to

15

cm

is

A

16J

B

8J

C

32J

D

24J

Correct Answer

Option B

Solution

When we extend the spring by

dx

then the work done

dW = k\,x\,dx

Applying integration both sides we get, $\therefore$

W = k\int\limits_{0.05}^{0.15} {x\,dx}

= {{800} \over 2}\left[ {{{\left( {0.15} \right)}^2} - {{\left( {0.05} \right)}^2}} \right]

= 8\,J

Q49

Four particles

A, B, C, D

of mass

\dfrac{m}{2}, m, 2 m, 4 m

, have same momentum, respectively. The particle with maximum kinetic energy is :

A B

B C

C D

D A

Correct Answer

Option D

Solution

The momentum $p$ of a particle is given by the product of its mass $m$ and its velocity $v$ , that is, $p = m \cdot v$ .

For a given momentum, the relationship between mass and velocity can be understood as inversely proportional.

This means that as the mass increases, the velocity decreases to maintain the same momentum, and vice versa.

The kinetic energy ( $K.E.$ ) of a particle is given by the formula $K.E. = \dfrac{1}{2} m v^2$ .

This equation shows that the kinetic energy depends on both the mass of the particle and the square of its velocity.

Given that four particles $A, B, C, D$ have masses $\dfrac{m}{2}, m, 2 m, 4 m$ , respectively, and all have the same momentum, we can assume the momentum of each particle to be $p$ .

This common value of momentum allows us to express the velocity of each particle in terms of its mass and the common momentum $p$ .

The velocity $v$ of each particle will be $v = \dfrac{p}{m}$ .

Thus, for each particle, we can determine the velocity as follows: For $A$ : $v_A = \dfrac{p}{\dfrac{m}{2}} = \dfrac{2p}{m}$ For $B$ : $v_B = \dfrac{p}{m}$ For $C$ : $v_C = \dfrac{p}{2m} = \dfrac{p}{2m}$ For $D$ : $v_D = \dfrac{p}{4m}$ Now, substituting these velocities into the kinetic energy formula yields the kinetic energies for each particle: $K.E._A = \dfrac{1}{2} \cdot \dfrac{m}{2} \cdot \left(\dfrac{2p}{m}\right)^2 = \dfrac{1}{2} \cdot \dfrac{m}{2} \cdot \dfrac{4p^2}{m^2} = \dfrac{2p^2}{m}$ $K.E._B = \dfrac{1}{2} \cdot m \cdot \left(\dfrac{p}{m}\right)^2 = \dfrac{1}{2} \cdot m \cdot \dfrac{p^2}{m^2} = \dfrac{p^2}{2m}$ $K.E._C = \dfrac{1}{2} \cdot 2m \cdot \left(\dfrac{p}{2m}\right)^2 = \dfrac{1}{2} \cdot 2m \cdot \dfrac{p^2}{4m^2} = \dfrac{p^2}{4m}$ $K.E._D = \dfrac{1}{2} \cdot 4m \cdot \left(\dfrac{p}{4m}\right)^2 = \dfrac{1}{2} \cdot 4m \cdot \dfrac{p^2}{16m^2} = \dfrac{p^2}{8m}$ Comparing these kinetic energies, we see that the particle $A$ has the maximum kinetic energy, as it is inversely related to mass in this scenario, and $A$ has the least mass but the highest velocity squared component, thus maximizing its kinetic energy.

Therefore, the correct answer is: Option D: A

Q50

A particle is projected at

60^\circ

to the horizontal with a kinetic energy K. The kinetic energy at the highest point is

A K/2

B K

C Zero

D K/4

Correct Answer

Option D

Solution

Let

u

be the velocity with which the particle is thrown and

m

be the mass of the particle. Then

KE = {1 \over 2}m{u^2}.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

At the highest point the velocity is

u

\cos \,{60^ \circ }

(only the horizontal component remains, the vertical component being zero at the top-most point).

Therefore kinetic energy at the highest point,

{\left( {KE} \right)_H} = {1 \over 2}m{u^2}{\cos ^2}60^\circ

\,\,\,\,\,\,\,\,\,\,\,\, = {K \over 4}

[ From eq

(1)

]