Work Power & Energy — Page 3 | NEET Physics

Q21

Two particles A and B, move with constant velocities

\overrightarrow {{v_1}}

and

\overrightarrow {{v_2}}

. At the initial moment their position vectors are

\overrightarrow {{r_1}}

and

\overrightarrow {{r_2}}

respectively. The condition for particles A and B for their collision is

A

{\overrightarrow r _1} \times {\overrightarrow v _1} = {\overrightarrow r _2} \times {\overrightarrow v _2}

B

{\overrightarrow r _1} - {\overrightarrow r _2} = {\overrightarrow v _1} - {\overrightarrow v _2}

C

{{{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} \over {\left| {{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} \right|}} = {{{{\overrightarrow v }_2} - {{\overrightarrow v }_1}} \over {\left| {{{\overrightarrow v }_2} - {{\overrightarrow v }_1}} \right|}}

D

{\overrightarrow r _1}.{\overrightarrow v _1} = {\overrightarrow r _2}.{\overrightarrow v _2}

Correct Answer

Option C

Solution

For collision

{\overrightarrow V _{B/A}}

should be along

\overline {B \to A} \left( {{{\overrightarrow r }_{A/B}}} \right)

So,

{{\overrightarrow {{V_2}} - \overrightarrow {{V_1}} } \over {\left| {\overrightarrow {{V_2}} - \overrightarrow {{V_1}} } \right|}} = {{\overrightarrow {{r_1}} - \overrightarrow {{r_2}} } \over {\left| {\overrightarrow {{r_1}} - \overrightarrow {{r_2}} } \right|}}

Q22

On a frictionless surface, a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle

\theta

to its initial direction and has a speed

{v \over 3}.

The second block's speed after the collision is

A

{3 \over {\sqrt 2 }}v

B

{{\sqrt 3 } \over 2}v

C

{{2\sqrt 2 } \over 3}v

D

{3 \over 4}v

Correct Answer

Option C

Solution

The situation is shown in the figure.

Let v' be speed of second block after the collision.

As the collision is elastic, so kinetic energy is conserved.

According to conservation of kinetic energy,

{1 \over 2}M{v^2} + 0 = {1 \over 2}M{\left( {{v \over 3}} \right)^2} + {1 \over 2}Mv{'^2}

{v^2} = {{{v^2}} \over 9} + v{'^2}

\Rightarrow v{'^2} = {v^2} - {{{v^2}} \over 9} = {{9{v^2} - {v^2}} \over 9} = {8 \over 9}{v^2}

v' = \sqrt {{8 \over 9}{v^2}} = {{\sqrt 8 } \over 3}v = {{2\sqrt 2 } \over 3}v

Q23

The heart of a man pumps 5 litres of blood through the arteries per minute at a pressure of 150 mm of mercury. If the density of mercury be 13.6

\times

10 3 kg/m 3 and g = 10 m/s 2 then the power (in watt) is

A 3.0

B 1.50

C 1.70

D 2.35

Correct Answer

Option C

Solution

Here, Volume of blood pumped by man’s heart, V = 5 litres = 5 × 10 –3 m 3 ( $\because$ 1 litre = 10–3 m 3 ) Time in which this volume of blood pumps, t = 1 min = 60 s Pressure at which the blood pumps, P = 150 mm of Hg = 0.15 m of Hg = (0.15 m)(13.6 × 10 3 kg/m 3 )(10 m/s 2 ) ( $\because$

P = h\rho g

) = 20.4 × 103 N/m 2 $\therefore$ Power of the heart =

{{PV} \over t}

= {{\left( {20.4 \times {{10}^3}N/{m^2}} \right)\left( {5 \times {{10}^{ - 3}}{m^3}} \right)} \over {60\,s}} = 1.70\,W

Q24

A block of mass 10 kg, moving in x direction with a constant speed of 10 m s

-

1 , is subjected to a retarding force F = 0.1x J/m during its travel from x = 20 m to 30 m. Its final KE will be

A 275 J

B 250 J

C 475 J

D 450 J

Correct Answer

Option C

Solution

Here, m = 10 kg, v i = 10 ms –1 Initial kinetic energy of the block is

{K_i} = {1 \over 2}mv_i^2 = {1 \over 2} \times \left( {10\,kg} \right) \times {\left( {10\,m{s^{ - 1}}} \right)^2} = 500\,J

Work done by retarding force

W = \int\limits_{{x_1}}^{{x_2}} {{F_r}dx = \int\limits_{20}^{30} { - 0.1\,xdx = - 0.1} \left[ {{{{x^2}} \over 2}} \right]} _{20}^{30}

= - 0.1\left[ {{{900 - 400} \over 2}} \right] = - 25\,J

According to work-energy theorem, W = K f – K i K f = W + K i = – 25 J + 500 J = 475 J

Q25

Two similar springs P and Q have spring constants K P and K Q , such that K P > K Q . They are stretched first by the same amount (case a), then by the same force (case b). The work done by the springs W P and W Q are related as, in case (a) and case (b) respectively

A W P > W Q ; W Q > W P

B W P < W Q ; W Q < W P

C W P = W Q ; W P > W Q

D W P = W Q ; W P = W Q

Correct Answer

Option A

Solution

Here, K P > K Q Case (a) : Elongation (x) in each spring is same.

{W_P} = {1 \over 2}{K_P}{x^2},{W_Q} = {1 \over 2}{K_Q}{x^2}

\therefore {W_P} > {W_Q}

Case (b) : Force of elongation is same. So,

{x_1} = {F \over {{K_P}}}

and

{x_2} = {F \over {{K_Q}}}

{W_P} = {1 \over 2}{K_P}x_1^2 = {1 \over 2}{{{F^2}} \over {{K_P}}}

{W_Q} = {1 \over 2}{K_Q}x_2^2 = {1 \over 2}{{{F^2}} \over {{K_Q}}}

\therefore {W_P} < {W_Q}

Q26

A particle of mass m is driven by a machine that delivers a constant power k watts. If the particle starts from rest the force on the particle at time t is

A

\sqrt {2mk} {\,t^{ - 1/2}}

B

{1 \over 2}\sqrt {mk} \,{t^{ - 1/2}}

C

\sqrt {{{mk} \over 2}} \,{t^{ - 1/2}}

D

\sqrt {mk} \,{t^{ - 1/2}}

Correct Answer

Option C

Solution

As we know power P =

{{dw} \over {dt}}

\Rightarrow w = Pt = {1 \over 2}m{V^2}

So,

v = \sqrt {{{2Pt} \over m}}

Hence, acceleration

a = {{dV} \over {dt}} = \sqrt {{{2P} \over m}} .{1 \over {2\sqrt t }}

Therefore, force on the particle at time ‘t’

= ma = \sqrt {{{2K{m^2}} \over m}} .{1 \over {2\sqrt t }} = \sqrt {{{Km} \over {2t}}} = \sqrt {{{mK} \over 2}} {t^{ - 1/2}}

Q27

A particle with total energy E is moving in a potential energy region U(x). Motion of the particle is restricted to the region when

A U(x) < E

B U(x) = 0

C U(x)

\le

E

D U(x) > E

Correct Answer

Option C

Solution

As the particle is moving in a potential energy region.

$\therefore$ Kinetic energy $\ge$ 0 And, total energy E = K.E.

+ P.E.

$\Rightarrow$ U(x) $\le$ E

Q28

One coolie takes 1 minute to raise a suitcase through a height of 2 m but the second coolie takes 30 s to raise the same suitcase to the same height. The powers of two coolies are in the ratio

A 1 : 3

B 2 : 1

C 3 : 1

D 1 : 2

Correct Answer

Option D

Solution

Power,

\left( p \right) = {{Work\,done} \over {Time\,taken}}

Here work done (= mgh) is same in both cases.

\therefore {{{P_1}} \over {{P_2}}} = {{{t_2}} \over {{t_1}}} = {{30\,s} \over {1\,\min }} = {{30\,s} \over {60\,s}} = {1 \over 2}

Q29

A uniform force of

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

newton acts on a particle of mass 2 kg. Hence the particle is displaced from position

\left( {2\widehat i + \widehat k} \right)

metre to position

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

metre. The work done by the force on the particle is

A 13 J

B 15 J

C 9 J

D 6 J

Correct Answer

Option C

Solution

Here,

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

Initial position,

\overrightarrow {{r_1}} = \left( {2\widehat i + \widehat k} \right)m

Final position,

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

Displacement,

\overrightarrow r = \overrightarrow {{r_2}} - \overrightarrow {{r_1}}

\overrightarrow r = \left( {4\widehat i + 3\widehat j - \widehat k} \right)m - \left( {2\widehat i + \widehat k} \right)m = 2\widehat i + 3\widehat j - 2\widehat k\,m

Work done,

W = \overrightarrow F .\overrightarrow r = \left( {3\widehat i + \widehat j} \right).\left( {2\widehat i + 3\widehat j - 2\widehat k} \right) = 6 + 3 = 9\,J

Q30

A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude P 0 . The instantaneous velocity of this car is proportional to

A t 2 P 0

B t 1/2

C t

-

1/2

D

{t \over {\sqrt m }}

Correct Answer

Option B

Solution

Constant power of car P 0 = F.V = ma.v

{P_0} = m{{dv} \over {dt}}.v

P 0 dt = mvdv Integrating

{P_0}.t = {{m{v^2}} \over 2}

v = \sqrt {{{2{P_0}t} \over m}}

$\because$ P 0 , m and 2 are constant $\therefore$

v \propto \sqrt t