Work Power & Energy — Page 6 | NEET Physics

Q51

A particle moves along a circle of radius

\left( {{{20} \over \pi }} \right)

m with constants tangential acceleration. If the velocity of vthe particle is 80 m/s at the end of the second revoluation after motion has begun, the tangential acceleration is

A 40 m/s 2

B 640

\pi

m/s 2

C 160

\pi

m/s 2

D 40

\pi

m/s 2

Correct Answer

Option A

Solution

W mg =

\Delta

K $\Rightarrow$ -mg

\ell

= ½ mv 2 – ½ mu 2 or, mv 2 = m(u 2 – 2g

\ell

] or,

v = \sqrt {{u^2} - 2g\ell } \widehat j

\overrightarrow u = u\widehat i

\therefore \overrightarrow v - \overrightarrow u = \sqrt {{u^2} - 2g\ell } \widehat j - u\widehat i

\therefore \left| {\overrightarrow v - \overrightarrow u } \right| = {\left[ {\left( {{u^2} - 2g\ell } \right) + {u^2}} \right]^{{1 \over 2}}}

= \sqrt {2\left( {{u^2} - g\ell } \right)}

Q52

When a long spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 10 cm, the potential energy stored in it will be

A U/5

B 5U

C 10U

D 25U

Correct Answer

Option D

Solution

If k be the spring constant, then

U = {1 \over 2} \times k \times {\left( 2 \right)^2} = 2k

{U_{final}} = {1 \over 2} \times k \times {\left( {10} \right)^2} = 50k

\Rightarrow {U \over {{U_{final}}}} = {{2k} \over {50k}} = {1 \over {25}}

\Rightarrow {U_{final}} = 25U

Q53

A particle is projected making an angle of 45 o with horizontal having kinetic energy K. The kinetic energy at highest point will be

A

{K \over {\sqrt 2 }}

B

{K \over 2}

C 2K

D K

Correct Answer

Option B

Solution

Kinetic energy of the ball = K and angle of projection ( $\theta$ ) = 45 o Velocity of the ball at the highest point = v cos $\theta$

= v\,\,\cos \,\,{45^o} = {v \over {\sqrt 2 }}

Therefore kinetic energy of the ball $\Rightarrow$

{1 \over 2}m \times \left( {{v \over {\sqrt 2 }}} \right) = {1 \over 4}m{v^2} = {K \over 2}

Q54

Two springs A and B having spring constant K A and K B (K A = 2K B ) are stretched by applying force of equal magnitude. If energy stored in spring A is E A then energy stored in B will be

A 2E A

B E A /4

C E A /2

D 4E A

Correct Answer

Option A

Solution

Energy =

{1 \over 2}K{x^2} = {1 \over 2}{{{F^2}} \over K}

{{{K_A}} \over {{K_B}}} = 2

$\therefore$

{{{E_A}} \over {{E_B}}} = {1 \over 2} \Rightarrow {E_B} = 2{E_A}

Q55

A child is sitting on a swing. Its minimum and maximum heights from the ground 0.75 m and 2 m respectively, its maximum speed will be

A 10 m/s

B 5 m/s

C 8 m/s

D 15 m/s

Correct Answer

Option B

Solution

Drop in P.E. = maximum K.E. mg(2 – 0.75) = 1mv 2 $\Rightarrow$ v =

\sqrt {2g\left( {1.25} \right)} = 5\,m/s

Q56

If

\overrightarrow F = \left( {60\widehat i + 15\widehat j - 3\widehat k} \right)

N and

\overrightarrow v = \left( {2\widehat i - 4\widehat j + 5\widehat k} \right)

m/s, then instantaneous power is

A 195 watt

B 45 watt

C 75 watt

D 100 watt.

Correct Answer

Option B

Solution

P =

\overrightarrow F .\overrightarrow v = \left( {60\widehat i + 15\widehat j - 3\widehat k} \right).\left( {2\widehat i - 4\widehat j + 5\widehat k} \right)

= 120 – 60 – 15 = 45 watts.

Q57

A mass of 1 kg is thrown up with a velocity of 100 m/s. After 5 seconds, it explodes into two parts. One part of mass 400 g comes down with a velocity 25 m/s. The velocity of other part is (Take g = 10 ms

-

2 )

A 40 m/s

B 50 m/s

C 100 m/s

D 60 m/s

Correct Answer

Option C

Solution

Velocity after 5 sec, v = u – gt = 100 – 10 × 5 = 50 m/s By conservation of momentum 1 × 50 = 0.4 × (–25) + 0.6 × v' 60 = 0.6 × v' $\Rightarrow$ v' = 100 m/s upwards

Q58

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: $\mu = \dfrac{\dfrac{1}{2} \cdot (0.1 \, \text{kg} + 3.9 \, \text{kg}) \cdot (10 \, \text{m/s})^2}{(0.1 \, \text{kg} + 3.9 \, \text{kg}) \cdot 10 \, \text{m/s}^2 \cdot 20 \, \text{m}} = 0.25$

Q59

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.

Q60

Sand is being dropped from a stationary dropper at a rate of

0.5 \,\mathrm{kgs}^{-1}

on a conveyor belt moving with a velocity of

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

. The power needed to keep the belt moving with the same velocity will be :

A 1.25 W

B 2.5 W

C 6.25 W

D 12.5 W

Correct Answer

Option D

Solution

{{dm} \over {dt}} = 0.5

kg/s

v = 5

m/s

F = {{vdm} \over {dt}} = 2.5

kg m/s2

P = \overline F \,.\,\overline v = (2.5)(5)

W

= 12.5

W