First slide

Elastic behaviour of solids

Question

When a 4 kg mass is hung vertically on a light spring that obeys Hooke's law, the spring stretches by 2 cm. The work required to be done by an external agent in stretching this spring by 5 cm will be (g =10 m/sec²)

Moderate

A

4.900 J
B

2.5 J
C

0.495 J
D

0.245 J

Solution

$K = \frac{F}{x} = \frac{40}{2 \times 10^{- 2}} = 2 \times 10^{3} N / m$
$Work done = \frac{1}{2} {Kx}^{2} = \frac{1}{2} \times 2000 \times (0 .05)^{2} = 2 .5 J$

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