Work Power & Energy — Page 2 | NEET Physics

Q11

A particle is released from height S from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively -

A

{S \over 4},\sqrt {{{3gS} \over 2}}

B

{S \over 4},{{3gS} \over 2}

C

{S \over 4},{{\sqrt {3gS} } \over 2}

D

{S \over 2},{{\sqrt {3gS} } \over 2}

Correct Answer

Option A

Solution

Take the surface of the Earth as the reference level for potential energy.

Initially, the particle is released from height $S$ , so Initial potential energy $= mgS$ Initial kinetic energy $= 0$ So, total mechanical energy is

E = mgS

Now suppose at some instant the particle is at height $h$ above the surface.

Then, Potential energy at that instant $= mgh$ Kinetic energy at that instant is given to be three times the potential energy:

K = 3U = 3mgh

Using conservation of mechanical energy:

K + U = mgS

Substitute $K = 3mgh$ and $U = mgh$ :

3mgh + mgh = mgS

4mgh = mgS

h = \frac{S}{4}

So the height from the surface is

\frac{S}{4}

Now find the speed at that instant. Since

K = \frac{1}{2}mv^2

and also

K = 3mgh

So,

\frac{1}{2}mv^2 = 3mgh

Cancel $m$ :

\frac{1}{2}v^2 = 3gh

v^2 = 6gh

Now put $h = \dfrac{S}{4}$ :

v^2 = 6g \cdot \frac{S}{4} = \frac{3gS}{2}

Therefore,

v = \sqrt{\frac{3gS}{2}}

Hence, the correct answer is:

\boxed{\frac{S}{4},\ \sqrt{\frac{3gS}{2}}}

So, Option A is correct.

Q12

Water falls from a height of 60m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine? (g = 10 m/s 2 )

A 7.0 kW

B 10.2 kW

C 8.1 kW

D 12.3 kW

Correct Answer

Option C

Solution

E = mgh P input =

{{mgh} \over t}

= {{15 \times 10 \times 60} \over 1} = 9000 = 9

kW 10% loss = 0.9 $\times$ 10 3 P output = 9 $\times$ 10 3 $-$ 0.9 $\times$ 10 3 = 8.1 kW

Q13

A force F = 20 + 10y acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is :

A 5 J

B 25 J

C 20 J

D 30 J

Correct Answer

Option B

Solution

Work done under the given variable force is : W =

\int\limits_{{y_1}}^{{y_1}} {Fdy}

Here, y 1 = 0, y 2 = 1 m $\therefore$ W =

\int\limits_0^1 {\left( {20 + 10y} \right)dy}

=

\left[ {20y + {{10{y^2}} \over 2}} \right]_0^1

= 25 J

Q14

A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to

A D

B

{3 \over 2}D

C

{7 \over 5}D

D

{5 \over 4}D

Correct Answer

Option D

Solution

Since from the given figure, in order to complete the vertical circle, velocity at the bottom of vertical circle should be

\sqrt {5gR}

.

As body is at rest initially, i.e., speed = 0.

At point A, speed = v.

As track is frictionless, so total mechanical energy will remain constant. $\therefore$ 0 + mgh =

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

+ 0 $\Rightarrow$ h =

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

For completing the vertical circle, v $\ge$

\sqrt {5gR}

$\therefore$ h =

{{{5gR} \over {2g}}}

=

{5 \over 2}R

=

{5 \over 4}D

Q15

Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m s

-

1 . Take 'g' constant with a value 10 m s

-

2 . The work done by the (i) gravitational force and the (ii) resistive force of air is

A (i) 1.25 j (ii)

-

8.25 J

B (i) 100 j (ii) 8.75 J

C (i) 10 j (ii)

-

8.75 J

D (i)

-

10 j (ii)

-

8.25 J

Correct Answer

Option C

Solution

From work-energy theorem, W g + W a =

\Delta

K.E or mgh + W a =

{1 \over 2}m{v^2} - 0

{10^{ - 3}} \times 10 \times {10^3} + {W_a} = {1 \over 2} \times {10^{ - 3}} \times {\left( {50} \right)^2}

\Rightarrow {W_a} = - 8.75\,J

which is the work done due to air resistance Work done due to gravity = mgh = 10 –3 × 10 × 10 3 = 10 J

Q16

A particle moves from a point

\left( { - 2\widehat i + 5\widehat j} \right)

to

\left( {4\widehat j + 3\widehat k} \right)

when a force of

\left( {4\widehat i + 3\widehat j} \right)

N is applied. How much work has been done by the force ?

A 8 J

B 11 J

C 5 J

D 2 J

Correct Answer

Option C

Solution

Here

\overrightarrow {{r_1}} = \left( { - 2\widehat i + 5\widehat j} \right)m

\overrightarrow {{r_2}} = \left( { - 4\widehat j + 3\widehat k} \right)m

\overrightarrow F = \left( {4\widehat i + 3\widehat j} \right)N

Work done by force F in moving from

\overrightarrow {{r_1}}

and

\overrightarrow {{r_2}}

W = \overrightarrow F \left( {\overrightarrow {{r_2}} - \overrightarrow {{r_1}} } \right)

\Rightarrow W = \left( {4\widehat i + 3\widehat j} \right)\left( {4\widehat j + 3\widehat k + 2\widehat i - 5\widehat j} \right)

\Rightarrow W = \left( {4\widehat i + 3\widehat j} \right).\left( {2\widehat i - \widehat j + 3\widehat k} \right)

\Rightarrow W = {\rm{ }}8{\rm{ }} + {\rm{ }}\left( {-3} \right){\rm{ }} = {\rm{ }}5{\rm{ }}J

Q17

What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?

A

\sqrt {3gR}

B

\sqrt {5gR}

C

\sqrt {gR}

D

\sqrt {2gR}

Correct Answer

Option B

Solution

To complete the loop a body must enter a vertical loop of radius R with the minimum velocity v =

\sqrt {5gR}

.

Q18

A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic enegy of the particle becomes equal to 8

\times

10

-

4 J by the end of the second revoluation after the beginning of the motion ?

A 0.18 m/s 2

B 0.2 m/s 2

C 0.1 m/s 2

D 0.15 m/s 2

Correct Answer

Option C

Solution

Given: Mass of particle, M = 10g

= {{10} \over {1000}}kg

radius of circle R = 6.4 cm Kinetic energy E of particle = 8 × 10 –4 J acceleration a t = ?

{1 \over 2}m{v^2} = E \Rightarrow {1 \over 2}\left( {{{10} \over {1000}}} \right){v^2} = 8 \times {10^{ - 4}}

\Rightarrow {v^2} = 16 \times {10^{ - 2}}

\Rightarrow v = 4 \times {10^{ - 1}} = 0.4\,m/s

Now, using v 2 = u 2 + 2a t s (s = 4 $\pi$ R)

{\left( {0.4} \right)^2} = {0^2} + 2{a_t}\left( {4 \times {{22} \over 7} \times {{6.4} \over {100}}} \right)

\Rightarrow {a_t} = {\left( {0.4} \right)^2} \times {{7 \times 100} \over {8 \times 22 \times 6.4}} = 0.1\,m/{s^2}

Q19

A body of mass 1 kg begins to move under the action of a time dependent force

\overrightarrow F = \left( {2t\widehat i + 3{t^2}\widehat J} \right)N,

where

{\widehat i}

and

{\widehat j}

are unit vectors along x and y axis. What power will be developed by the force at the time t ?

A (2t 3 + 3t 4 ) W

B (2t 3 + 3t 5 ) W

C (2t 2 + 3t 3 ) W

D (2t 2 + 4t 4 ) W

Correct Answer

Option B

Solution

Given force

(\overrightarrow F) = 2t\widehat i + 3{t^2}\widehat j

According to Newton's second law of motion,

m{{d\overrightarrow v } \over {dt}} = 2t\widehat i + 3{t^2}\widehat j\,\,(m = 1kg)

\Rightarrow \int\limits_0^{\overrightarrow v } {d\overrightarrow v } = \int\limits_0^t {\left( {2t\widehat i + 3{t^2}\widehat j} \right)} dt

\Rightarrow v = {t^2}\widehat i + {t^3}\widehat j

Power (P) =

\overrightarrow F .\overrightarrow v \left( {2t\widehat i + 3{t^2}\widehat j} \right)\left( {{t^2}\widehat i + {t^3}\widehat j} \right)

= (2t 3 + 3t 5 )W

Q20

A ball is thrown vertically downwards from a height of 20 m with an initial velocity v 0 . It collides with the ground, losses 50 percent of its energy in collision and rebounds to the same height. The initial velocity v 0 is (Take g = 10 m s

-

2 )

A 28 m s

-

1

B 10 m s

-

1

C 14 m s

-

1

D 20 m s

-

1

Correct Answer

Option D

Solution

When ball collides with the ground it loses its 50% of energy

{{K.{E_f}} \over {K.{E_i}}} = {1 \over 2} \Rightarrow {{{1 \over 2}mV_f^2} \over {{1 \over 2}mV_i^2}} = {1 \over 2}

\Rightarrow {{{V_f}} \over {{V_i}}} = {1 \over {\sqrt 2 }}

\Rightarrow {{\sqrt {2gh} } \over {\sqrt {V_0^2 + 2gh} }} = {1 \over {\sqrt 2 }}

or

4gh = V_0^2 + 2gh

$\therefore$ V 0 = 20 ms –1