Work Power & Energy — Page 7 | JEE Physics

Q61

A body of mass m starts moving from rest along x-axis so that its velocity varies as

\upsilon = a\sqrt s

where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first t seconds after the start of the motion is :

A

{1 \over 8}\,

m a4 t2

B 8 m a4 t2

C 4 m a4 t2

D

{1 \over 4}\,

m a4 t2

Correct Answer

Option A

Solution

Given,

\upsilon

= a

\sqrt s

$\Rightarrow$

\,\,\,

{{ds} \over {dt}} = a\sqrt s

$\Rightarrow$

\,\,\,

\int\limits_0^t {{{ds} \over {\sqrt s }}} = \int\limits_0^z {a\,dt}

$\Rightarrow$

\,\,\,

2

\sqrt s

= at $\Rightarrow$

\,\,\,

s =

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

=

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

\therefore\,\,\,\,

acceleration =

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

\therefore\,\,\,

Force (F) = m $\times$

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

\therefore\,\,\,\,

Work done = F. S =

{{m{a^2}} \over 2} \times {{{a^2}{t^2}} \over 4}

=

{{m{a^4}{t^2}} \over 8}

Q62

Two particles of the same mass m are moving in circular orbits because of force, given by

F\left( r \right) = {{ - 16} \over r} - {r^3}

The first particle is at a distance r = 1, and the second, at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to :

A

6 \times {10^{ - 2}}

B

3 \times {10^{ - 3}}

C

{10^{ - 1}}

D

6 \times {10^{ 2}}

Correct Answer

Option A

Solution

In circular motion the force required

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

\therefore\,\,\,

{{m{v^2}} \over r} = {{16} \over r} + {r^3}

$\Rightarrow$ mv2 = 16 + r4

\therefore\,\,\,

kinetic energy (K) =

{1 \over 2}

mv2 =

{1 \over 2}

[ 16 + r4]

\therefore\,\,\,

Kinetic energy of first particle (K1) =

{1 \over 2}

[16 + 1] Kinetic energy of second particle (K2) =

{1 \over 2}

[16 + 44]

\therefore\,\,\,\,

{{{K_1}} \over {{K_2}}}

=

{{{{16 + 1} \over 2}} \over {{{16 + 256} \over 2}}}

=

{{17} \over {272}}

$\Rightarrow$

\,\,\,

{{{K_1}} \over {{K_2}}} = 6 \times {10^{ - 2}}

Q63

A particle of mass

m

moves on a straight line with its velocity increasing with distance according to the equation

v=\alpha \sqrt{x}

, where

\alpha

is a constant. The total work done by all the forces applied on the particle during its displacement from

x=0

to

x=\mathrm{d}

, will be :

A

\dfrac{\mathrm{m}}{2 \alpha^2 \mathrm{~d}}

B

\dfrac{\mathrm{md}}{2 \alpha^2}

C

\dfrac{\mathrm{m} \alpha^2 \mathrm{~d}}{2}

D

2 \mathrm{~m} \alpha^2 \mathrm{~d}

Correct Answer

Option C

Solution

To find the total work done by all forces applied on the particle during its displacement, we can use the work-energy theorem which states that the work done by all forces on an object is equal to the change in kinetic energy of the object.

So, we first need to find the initial and final kinetic energies of the particle and then calculate the work done.

The velocity of the particle is given by

v = \alpha \sqrt{x}

, and the kinetic energy

K

of the particle is given by

K = \frac{1}{2} m v^2

. We can substitute the expression for

v

into this formula to get the kinetic energy as a function of position

x

:

K(x) = \frac{1}{2} m (\alpha \sqrt{x})^2 = \frac{1}{2} m \alpha^2 x

To find the total work done from

x = 0

to

x = d

, we need to compute the difference in kinetic energy between these two points:

W = K(d) - K(0)

At

x = d

,

K(d) = \frac{1}{2} m \alpha^2 d

At

x = 0

, since the particle starts from this position,

K(0) = \frac{1}{2} m \alpha^2 (0) = 0

So, the work done

W

is simply the kinetic energy at

x = d

,

W = \frac{1}{2} m \alpha^2 d - 0 = \frac{1}{2} m \alpha^2 d

This matches with Option C:

\frac{m \alpha^2 d}{2}

.

Q64

A block moving horizontally on a smooth surface with a speed of 40 m/s splits into two parts with masses in the ratio of 1 : 2. If the smaller part moves at 60 m/s in the same direction, then the fractional change in kinetic energy is :-

A

{{1 \over 3}}

B

{{2 \over 3}}

C

{{1 \over 8}}

D

{{1 \over 4}}

Correct Answer

Option C

Solution

3MV0 = 2MV2 + MV1 3V0 = 2V2 + V1 120 = 2V2 + 60 $\Rightarrow$ V2 = 30 m/s

{{\Delta K.E.} \over {K.E.}} = {{{1 \over 2}MV_1^2 + {1 \over 2}2MV_2^2 - {1 \over 2}3MV_0^2} \over {{1 \over 2}3MV_0^2}}

= {{V_1^2 + 2V_2^2 - 3V_0^2} \over {3V_0^2}}

= {{3600 + 1800 - 4800} \over {4800}} = {1 \over 8}

Q65

A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial speed is v0 = 10 ms–1. If, after 10 s, its energy is

{1 \over 8}mv_0^2

, the value of k will be:

A 10-1 kg m-1 s-1

B 10-3 kg m-1

C 10-3 kg s-1

D 10-4 kg m-1

Correct Answer

Option D

Solution

According to the question, final kinetic energy =

{1 \over 8}mv_0^2

Let final speed of the body = Vf So final kinetic energy =

{1 \over 2}mv_f^2

According to question,

{1 \over 2}mv_f^2

=

{1 \over 8}mv_0^2

\Rightarrow {v_f} = {{{v_0}} \over 2}

=

{{10} \over 2}

= 5 m/s Given that, F = –kv2 $\Rightarrow$

m\left( {{{dv} \over {dt}}} \right)

= - k{v^2}

\Rightarrow {10^{ - 2}}\left( {{{dv} \over {dt}}} \right) = - k{v^2}

\Rightarrow \int\limits_{10}^5 {{{dv} \over {{v^2}}}} = - 100k\int\limits_0^{10} {dt}

\Rightarrow {1 \over 5} - {1 \over {10}} = 100k \times 10

\Rightarrow k = {10^{ - 4}}kg\,{m^{ - 1}}

Q66

A body of mass

2 \mathrm{~kg}

begins to move under the action of a time dependent force given by

\vec{F}=\left(6 t \hat{i}+6 t^2 \hat{j}\right) N

. The power developed by the force at the time

t

is given by:

A

\left(3 t^3+6 t^5\right) W

B

\left(9 t^5+6 t^3\right) W

C

\left(6 t^4+9 t^5\right) W

D

\left(9 t^3+6 t^5\right) W

Correct Answer

Option D

Solution

\begin{aligned} & \vec{F}=\left(6 t \hat{i}+6 t^2 \hat{j}\right) N \\ & \vec{F}=m \vec{a}=\left(6 t \hat{i}+6 t^2 \hat{j}\right) \\ & \vec{a}=\frac{\vec{F}}{m}=\left(3 t \hat{i}+3 t^2 \hat{j}\right) \\ & \vec{v}=\int\limits_0^t \vec{a} d t=\frac{3 t^2}{2} \hat{i}+t^3 \hat{j} \\ & P=\vec{F} \cdot \vec{v}=\left(9 t^3+6 t^5\right) W \end{aligned}

Q67

A particle experiences a variable force

\overrightarrow F = \left( {4x\widehat i + 3{y^2}\widehat j} \right)

in a horizontal x-y plane. Assume distance in meters and force is newton. If the particle moves from point (1, 2) to point (2, 3) in the x-y plane, then Kinetic Energy changes by :

A 50.0 J

B 12.5 J

C 25.0 J

D 0 J

Correct Answer

Option C

Solution

W = \int {\overrightarrow F \,.\,d\overrightarrow r }

= \int\limits_1^2 {4xdx + \int\limits_2^3 {3{y^2}dy} }

= [2{x^2}]_1^2 + [{y^3}]_2^3

= 2 \times 3 + (27 - 8)

= 25

J

Q68

A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement

x

is proportional to

A

x

B

{e^x}

C

{x^2}

D

{\log _e}x

Correct Answer

Option C

Solution

Given that, retardation $\propto$ displacement $\Rightarrow$

a=-kx

But we know

a = v{{dv} \over {dx}}\,\,\,\,\,\,\,\,\,

$\therefore$

{{vdv} \over {dx}} = - kx

\Rightarrow \int\limits_{{v_1}}^{{v_2}} v \,dv = - k\int\limits_0^x {xdx}

\left( {v_2^2 - v_1^2} \right) = - k{{{x^2}} \over 2}

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

$\therefore$ Loss in kinetic energy is proportional to

{x^2}

. $\therefore$

\Delta K \propto {x^2}

Q69

An elevator in a building can carry a maximum of 10 persons, with the average mass of each person being 68 kg, The mass of the elevator itself is 920 kg and it moves with a constant speed of 3 m/s. The frictional force opposing the motion is 6000 N. If the elevator is moving up with its full capacity, the power delivered by the motor to the elevator (g = 10 m/s2) must be at least :

A 48000 W

B 62360 W

C 56300 W

D 66000 W

Correct Answer

Option D

Solution

Net force on motor will be Fm = [920 + 68(10)]g + 6000 = 22000 N So, required power for motor P = Fm.V = 22000 $\times$ 3 = 66000 W

Q70

An object is dropped from a height h from the ground. Every time it hits the ground it looses 50% of its kinetic energy. The total distance covered as t

\to

\infty

is :

A 3h

B

\infty

C

{5 \over 3}

h

D

{8 \over 3}

h

Correct Answer

Option A

Solution

Let, Kinetic energy (k) =

{1 \over 2}

m

\upsilon

2 before it hit the ground. After hitting the ground kinetic energy (k') =

{1 \over 2}

m

\upsilon

_1^2

\therefore\,\,\,

According to the question,

{1 \over 2}

m

\upsilon

_1^2

=

{1 \over 2}

$\times$

{1 \over 2}

m

\upsilon

2 $\Rightarrow$

\,\,\,

\upsilon

1 =

{v \over {\sqrt 2 }}

After hitting the ground the object will bounce h' =

{{v_1^2} \over {2g}}

=

{{{v^2}} \over {4g}}

=

{h \over 2}

[ as h =

{{{v^2}} \over {2g}}

] Total distance travelled from the time it first hits the ground to the next time it hits the ground is =

{h \over 2}

+

{h \over 2}

= h So, this will create a infinite geometric progression with the common ration

{1 \over 2}

.

\therefore\,\,\,

Total distance covered = h (distance travelled by the obhect when first dropped, before it hits the ground) + (h +

{h \over 2}

+

{h \over 4}

+ . . . . . . . . $\propto$ ) = h +

{h \over {1 - {1 \over 2}}}

= h + 2h = 3h