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Work Energy And Power NEET Questions — Free MCQ Practice

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A block of mass 10 kg, moving in the x-direction with a constant speed of 10 ms^-1, is subjected to a retarding force F=0.1x J/m during its travel from x=20m to 30m. Its final KE will be :

A.

475 J

B.

450 J

C.

275 J

D.

250 J

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Explanation

_(1)
=>Kf_{=W+Ki= $From work - energy theorem : Work done = Change in K . E . \Rightarrow W = K_{f} - K_{i} \Rightarrow K_{f} = W + K_{i} = \int_{x_{1}}^{x_{2}} F_{x} dx + \frac{1}{2} {mv}^{2} \Rightarrow K_{f} = - \int_{20}^{30} 0.1 xdx + \frac{1}{2} \times 10 \times 10^{2} = 500 - 0.1 {[\frac{x^{2}}{2}]}_{20}^{30} \Rightarrow K_{f} = 475 J$}

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

A.

$\sqrt{(\frac{m k}{2})} t^{- 1 / 2}$

B.

$\sqrt{(m k)} t^{- 1 / 2}$

C.

$\sqrt{(2 m k)} t^{- 1 / 2}$

D.

$\frac{1}{2} \sqrt{(m k)} t^{- 1 / 2}$

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Explanation

As the machine delivers a constant power, so F.v=constant=k(watts)
=> m $\frac{dv}{dt}$ .v=k

=> $\int$ vdv= $\frac{k}{m}$ $\int$ dt

=> v²/2= $\frac{k t}{m}$

=>v= $\sqrt{\frac{2 k t}{m}}$

Now, the force on the particle is given by

F=m $\frac{dv}{dt}$

= $\sqrt{(\frac{m k}{2})} t^{- 1 / 2}$

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Two particles of masses m₁,m₂ move with initial velocities u₁ and u₂. On collision, one of the particles get excited to higher level, after absorbing energy $ε$ . If final velocities of particles be v₁ and v₂, then we must have

A.

m₁²u₁+m₂²u₂- $ε$ =m₁²v₁+m₂²v₂

B.

$\frac{1}{2}$ m₁u₁²+ $\frac{1}{2}$ m₂u₂= $\frac{1}{2}$ m₁v₁²+ $\frac{1}{2}$ m₂v₂²- $ε$

C.

$\frac{1}{2}$ m₁u₁²+ $\frac{1}{2}$ m₂u₂²- $ε$ = $\frac{1}{2}$ m₁v₁²+ $\frac{1}{2}$ m₂v₂²

D.

$\frac{1}{2}$ m₁²u₁²+ $\frac{1}{2}$ m₂²u₂²+ $ε$ = $\frac{1}{2}$ m₁²v₁²+ $\frac{1}{2}$ m₂²v₂²

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Explanation

Total initial energy= $\frac{1}{2}$ m₁u₁²+ $\frac{1}{2}$ m₂u₂²

Since, after collision one particle absorb energy $ε$ .

∴ Total final energy= $\frac{1}{2}$ m₁v₁²+ $\frac{1}{2}$ m₂v₁²+ $ε$

From conservation of energy,

$\frac{1}{2}$ m₁u₁²+ $\frac{1}{2}$ m₂u_2²= $\frac{1}{2}$ m₁v₁²+ $\frac{1}{2}$ m₂v₂²+ $ε$

=> $\frac{1}{2}$ m₁u₁²+ $\frac{1}{2}$ m₂u₂²- $ε$ = $\frac{1}{2}$ m₁v₁²+ $\frac{1}{2}$ m₂v₂²

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A uniform force of (3i + j) N acts on a particle of mass 2 kg. Hence the particle is displaced from position (2i+k) m to position (4i+3j-k) m. The work done by the force on the particle is-

A.

9J

B.

6J

C.

13J

D.

15J

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Explanation

(a) Given, force F=3i+j

∴ r₁=(2i+k)m and r₂=(4i+3j-k)m

∴ s=r₂-r₁=(4i+3j-k)-(2i+k)=(2i+3j-2k)m

W=F $.$ s=(3i+j) $.$ (2i+3j-2k)

=3x2+3+0

=6+3=9J

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A body of mass m is taken from the earth’s surface to the height equal to twice the radius (R) of the earth. The change is potential energy of body will be

A.

mg2R

B.

2/3mgR

C.

3mgR

D.

1/3mgR

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Explanation

(b) Change in potential energy:

ΔU = $= - \frac{G M m}{R + 2 r} - (- \frac{G M m}{R}) = \frac{2 G M m}{3 R} = \frac{2}{3} m g R$

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The potential energy of a particle in a force field is $U = \frac{A}{r^{2}} - \frac{B}{r},$ where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is

A.

B/2A

B.

2A/B

C.

A/B

D.

B/A

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Explanation

Given, the potential energy of a particle in a force field $U = \frac{A}{r^{2}} - \frac{B}{r^{1}}$

For stable equilibrium, F= $- \frac{d U}{d r} = 0$

0=- $\frac{2 A}{r^{3}} + \frac{B}{r^{2}}$

or $\frac{2 A}{r} = B$

The distance of particle from the centre of the field

$r = \frac{2 A}{B}$

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The potential energy of a system increases if work is done

A.

by the system against a conservative force

B.

by the system against a nonconservative force

C.

upon the system by a conservative force

D.

upon the system by a nonconservative force

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Explanation

The potential energy of a system increases if work is done by the system against a conservative force.

$∆ U = - W_{conservative force} = W_{system}$

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A ball moving with velocity $2 m s^{- 1}$ collides head on with another stationery ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in $m s^{- 1}$ ) after collision will be

A.

0,1

B.

1,1

C.

1,0.5

D.

0,2

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Explanation

If two bodies collide head on with coefficient of restitution

$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$ ...(i)

From the law of conservation of linear momentum

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{1} v_{2}$

$\Rightarrow v_{1} = [\frac{m_{1} - e m_{2}}{m_{1} + m_{2}}] u_{1} + [\frac{(1 + e) m_{2}}{m_{1} + m_{2}}] u_{2}$

Substituting $u_{1} = 2 m s^{- 1}, u_{2} = 0, m_{1} = m a n d m_{2} = 2 m, e = 0.5$

we get $v_{1} = [\frac{m - m}{m + 2 m}] \times 2$

$\Rightarrow v_{1} = 0$

Similarly,

$v_{2} = [\frac{(1 + e) m_{1}}{m_{1} + m_{2}}] u_{1} + [\frac{m_{2} - e m_{1}}{m_{1} + m_{2}}] u_{2}$

$= [\frac{1.5 \times m}{3 m}] \times 2$

= $1 m s^{^{- 1}}$

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An engine pumps water through a hosepipe. Water passes through the pipe and leaves it with a velocity of 2 $m s^{- 1} .$ The mass per unit length of water in the pipe is 100 $k g m^{- 1} .$ What is the power of the engine?

A.

400W

B.

200W

C.

100W

D.

800W

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Explanation

$F = \frac{m (v - u)}{t} = \frac{mv}{t} if u = 0 . P = F . v = \frac{{mv}^{2}}{t} As \frac{m}{t} = linear density \times velocity = μv P = {μv}^{3} = 800 W$

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A particle of mass M starting from rest undergoes uniform acceleration. If the speed acquired in time T is v, the power delivered to the particle is

A.

$\frac{{Mv}^{2}}{T}$

B.

$\frac{1}{2} \frac{{Mv}^{2}}{T^{2}}$

C.

$\frac{{Mv}^{2}}{T^{2}}$

D.

$\frac{1}{2} \frac{{Mv}^{2}}{T}$

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Explanation

$The kinetic energy of particle = \frac{1}{2} {Mv}^{2} Power = \frac{Energy}{Time} P = \frac{1}{2} \frac{{Mv}^{2}}{T}$

Work Energy And Power MCQs for NEET — Physics Questions with Answers

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