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Work Energy And Power NEET Questions — Free MCQ Practice

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An ice cream has a marked value of 700 kcal. How many kilowatt- hour of energy will it deliver to the body as it is digested

A.

0.81 kWh

B.

0.90 kWh

C.

1.11 kWh

D.

0.71 kWh

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Explanation

It is based on conversion of units from kcal to KWh

$1 k c a l = 10^{3} C a l o r i e = 4200 J$ .................. (1)

$1 k W h = 1 k W X 1 h = 10^{3} \frac{J}{s} \times 3600 s = 3.6 \times 10^{6} J 1 J = \frac{1 k W h}{3.6 \times 10^{6}} ∴ 4200 J = \frac{4200 k W h}{3.6 \times 10^{6}} . . . . (2) F r o m (1) a n d (2), 1 k c a l = \frac{4200 k W h}{3.6 \times 10^{6}} 700 k c a l = \frac{700 \times 4200 k W h}{3.6 \times 10^{6}} = 0.81 k W h$

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A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by 1m/s so as to have same K.E. as that of the boy. The original speed of the man will be

A.

$\sqrt{2} m / s$

B.

$(\sqrt{2} - 1) m / s$

C.

$\frac{1}{(\sqrt{2} - 1)} m / s$

D.

$\frac{1}{\sqrt{2}} m / s$

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Explanation

$L e t, m a s s o f m a n = m t h e n, m a s s o f b o y = \frac{m}{2} v = v e l o c i t y o f b o y, V = v e l o c i t y o f m a n$

$K E o f t h e m a n = \frac{1}{2} K E o f t h e b o y \frac{1}{2} m V^{2} = \frac{1}{2} [\frac{1}{2} \frac{m}{2} v^{2}] - - - - - - - - - - (i)$

$W h e n m a n i n c r e a s e s h i s s p e e d b y 1 m / s t h e n, K E o f t h e m a n = K E o f t h e b o y \frac{1}{2} m {(V + 1)}^{2} = [\frac{1}{2} \frac{m}{2} v^{2}] - - - - - - - - - - - (i i)$

Solve both equations, $V = \frac{1}{\sqrt{2} - 1}$

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A particle of mass m at rest is acted upon by a force F for a time t. Its Kinetic energy after an interval t is

A.

$\frac{F^{2} t^{2}}{m}$

B.

$\frac{F^{2} t^{2}}{2 m}$

C.

$\frac{F^{2} t^{2}}{3 m}$

D.

$\frac{F t}{2 m}$

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Explanation

Kinetic energy $E = \frac{P^{2}}{2 m} = \frac{{(F t)}^{2}}{2 m} = \frac{F^{2} t^{2}}{2 m}$ [As P = F t]

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Two bodies of masses m and 4 m are moving with equal K.E. The ratio of their linear momentums is

A.

4 : 1

B.

1 : 1

C.

1 : 2

D.

1 : 4

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Explanation

$p = \sqrt{2 m E}$

∴ $P \propto \sqrt{m}$

⇒ $\frac{P_{1}}{P_{2}} = \sqrt{\frac{m_{1}}{m_{2}}} = \sqrt{\frac{m}{4 m}} = \frac{1}{2}$

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A particle of mass m₁ is moving with a velocity v₁ and another particle of mass m₂ is moving with a velocity v₂. Both of them have the same momentum but their different kinetic energies are E₁ and E₂ respectively. If m₁ > m₂ then

A.

E₁ < E₂

B.

$\frac{E_{1}}{E_{2}} = \frac{m_{1}}{m_{2}}$

C.

E₁ > E₂

D.

E₁ = E₂

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Explanation

$E = \frac{P^{2}}{2 m}$ if P = constant then $E \propto \frac{1}{m}$

According to problem $m_{1} > m_{2}$ ∴ $E_{1} < E_{2}$

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Four particles are given, having same momentum. Which one has maximum kinetic energy-

A.

Proton

B.

Electron

C.

Deutron

D.

α-particles

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Explanation

$E = \frac{P^{2}}{2 m}$

∴ $E \propto \frac{1}{m}$ (If P = constant)

i.e. the lightest particle will possess maximum kinetic energy and in the given option mass of electron is minimum.

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If a man increase his speed by 2 m/s , his K.E. is doubled, the original speed of the man is

A.

$(1 + 2 \sqrt{2}) m / s$

B.

4 m/s

C.

$(2 + 2 \sqrt{2}) m / s$

D.

$(2 + \sqrt{2}) m / s$

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Explanation

Initial kinetic energy $E = \frac{1}{2} m v^{2}$ …(i)

Final kinetic energy $2 E = \frac{1}{2} m {(v + 2)}^{2}$ …(ii)

by solving equation (i) and (ii) we get

$v = (2 + 2 \sqrt{2}) m / s$

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A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It slides down a smooth surface to the ground, then climbs up another hill of height 30 m and finally slides down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is

A.

10 m/s

B.

$10 \sqrt{30}$ m/s

C.

40 m/s

D.

20 m/s

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A body of mass m accelerates uniformly from rest to v₁ in time t₁. As a function of time t, the instantaneous power delivered to the body is

A.

$\frac{m v_{1} t}{t_{1}}$

B.

$\frac{m v_{1}^{2} t}{t_{1}}$

C.

$\frac{m v_{1} t^{2}}{t_{1}}$

D.

$\frac{m v_{1}^{2} t}{t_{1}^{2}}$

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Explanation

$P = \vec{F} . \vec{v} = m a \times a t = m a^{2} t$ [as u = 0]

$= m {(\frac{v_{1}}{t_{1}})}^{2} t = \frac{m v_{1}^{2} t}{t_{1}^{2}}$ $[As a = v_{1} / t_{1}]$

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A weight lifter lifts 300 kg from the ground to a height of 2 meter in 3 second. The average power generated by him is

A.

5880 watt

B.

4410 watt

C.

2205 watt

D.

1960 watt

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Explanation

P $= \frac{Workdone}{Time}$ = $\frac{m g h}{t} = \frac{300 \times 9 .8 \times 2}{3} = 1960 W$

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

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