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The potential energy of a certain spring when stretched through a distance ‘S’ is 10 joule. The amount of work (in joule) that must be done on this spring to stretch it through an additional distance ‘S’ will be:

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Explanation

$\frac{1}{2} k S^{2} = 10 J$ (given in the problem)

$W = \frac{1}{2} k [{(2 S)}^{2} - {(S)}^{2}] = 3 \times \frac{1}{2} k S^{2}$ = 3 × 10 = 30 J

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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

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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

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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

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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

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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

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Explanation

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

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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

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Explanation

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

$= 18 .75 J$

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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

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Explanation

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

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

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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

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Explanation

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

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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

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Explanation

$U \propto x^{2}$

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

∴ $U_{2} = 25 U$

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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)

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

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

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

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

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Explanation

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

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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

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Explanation

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

⇒ $F = - \frac{d U}{d x} = 2 B x$

⇒ $F \propto x$

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