Blog Log in Get Started Free

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

Practice free Work Energy And Power (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Clear Register free for difficulty & keyword filters

● ● ● ● ●

A position dependent force $F = 7 - 2 x + 3 x^{2} n e w t o n$ acts on a small body of mass 2 kg and displaces it from x = 0 to x = 5 m. The work done in joules is:

270

135

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$W = \int_{0}^{5} F d x = \int_{0}^{5} (7 - 2 x + 3 x^{2}) d x$ = ${[7 x - x^{2} + x^{3}]}_{0}^{5}$

= 35 – 25 + 125 = 135 J

● ● ● ● ●

A body of mass 3 kg is under a force, which causes a displacement in it, given by $S = \frac{t^{3}}{3}$ (in m). Find the work done by the force in first 2 seconds:

2 J

3.8 J

5.2 J

24 J

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$S = \frac{t^{3}}{3}$ ∴ $d S = t^{2} d t$

$a = \frac{d^{2} S}{d t^{2}} = \frac{d^{2}}{d t^{2}} [\frac{t^{3}}{3}] = 2 t m / s^{2}$

Now work done by the force $W = \int_{0}^{2} F . d S = \int_{0}^{2} m a . d S$

$\int_{0}^{2} 3 \times 2 t \times t^{2} d t = \int_{0}^{2} 6 t^{3} d t$ = $\frac{3}{2} {[t^{4}]}_{0}^{2}$ = 24 J

● ● ● ● ●

A spring of force constant 800 N/m has an extension of 5cm. The work done in extending it from 5cm to 15 cm is:

16 J

8 J

32 J

24 J

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$W = \frac{1}{2} k (x_{2}^{2} - x_{1}^{2}) = \frac{1}{2} \times 800 \times (15^{2} - 5^{2}) \times 10^{- 4}$ $= 8 J$

● ● ● ● ●

A spring of spring constant 5 × 10³ N/m is stretched initially by 5cm from the unstretched position. Then the work required to stretch it further by another 5 cm is

6.25 N-m

12.50 N-m

18.75 N-m

25.00 N-m

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$W = \frac{1}{2} k (x_{2}^{2} - x_{1}^{2}) = \frac{1}{2} \times 5 \times 10^{3} (10^{2} - 5^{2}) \times 10^{- 4}$

$= 18 .75 J$

● ● ● ● ●

A mass of 0.5kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides with a nearly weightless spring of force constant k = 50 N/m. The maximum compression of the spring would be

0.15 m

0.12 m

1.5 m

0.5 m

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

The kinetic energy of mass is converted into potential energy of a spring

$\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$ ⇒ $x = \sqrt{\frac{m v^{2}}{k}} = \sqrt{\frac{0 .5 \times {(1 .5)}^{2}}{50}} = 0 .15 m$

● ● ● ● ●

A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to

x²

e^x

log_e x

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$According to the question : a = - k x \int_{u}^{v} v d v = - \int_{0}^{x} k x d x \frac{v^{2} - u^{2}}{2} = - \frac{k x^{2}}{2} \frac{m (v^{2} - u^{2})}{2} = - m \frac{k x^{2}}{2}$

● ● ● ● ●

If a long spring is stretched by 0.02 m, its potential energy is U. If the spring is stretched by 0.1 m, then its potential energy will be

$\frac{U}{5}$

25U

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$U \propto x^{2}$

⇒ $\frac{U_{2}}{U_{1}} = {(\frac{x_{2}}{x_{1}})}^{2} = {(\frac{0 .1}{0 .02})}^{2} = 25$

∴ $U_{2} = 25 U$

● ● ● ● ●

The spring extends by x on loading, then energy stored by the spring is : (if T is the tension in spring and k is spring constant)

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

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

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

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$x = \frac{T}{k} U = \frac{1}{2} k x^{2} = \frac{T^{2}}{2 k}$

● ● ● ● ●

The potential energy of a body is given by, U = A – Bx² (Where x is the displacement). The magnitude of force acting on the particle is

Constant

Proportional to x

Proportional to x²

Inversely proportional to x

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$U = A - B x^{2}$

⇒ $F = - \frac{d U}{d x} = 2 B x$

⇒ $F \propto x$

● ● ● ● ●

The potential energy between two atoms in a molecule is given by $U (x) = \frac{a}{x^{12}} - \frac{b}{x^{6}}$ ; where a and b are positive constants and x is the distance between the atoms. The atoms are in stable equilibrium when:

$x = \sqrt[6]{\frac{11 a}{5 b}}$

$x = \sqrt[6]{\frac{a}{2 b}}$

x = 0

$x = \sqrt[6]{\frac{2 a}{b}}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Condition for stable equilibrium $F = - \frac{d U}{d x} = 0$

⇒ $- \frac{d}{d x} [\frac{a}{x^{12}} - \frac{b}{x^{6}}] = 0$

⇒ $- 12 a x^{- 13} + 6 b x^{- 7} = 0$

⇒ $\frac{12 a}{x^{13}} = \frac{6 b}{x^{7}}$

⇒ $\frac{2 a}{b} = x^{6}$

⇒ $x = \sqrt[6]{\frac{2 a}{b}}$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Work Energy And Power question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.