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5.Work, Energy, Power and Collision
medium
When a $1.0\,kg$ mass hangs attached to a spring of length $50 cm$, the spring stretches by $2 \,cm$. The mass is pulled down until the length of the spring becomes $60\, cm.$ What is the amount of elastic energy stored in the spring in this condition, if $g = 10 m/s^{2}$ ............. $\mathrm{Joule}$
A
$1.5$
B
$2$
C
$2.5$
D
$3$
Solution
(c)Force constant of a spring
$k = \frac{F}{x} = \frac{{mg}}{x} = \frac{{1 \times 10}}{{2 \times {{10}^{ – 2}}}}$
$⇒$ $k = 500\,N/m$
Increment in the length $= 60 -50 = 10 cm$
$U = \frac{1}{2}k{x^2} = \frac{1}{2}500\,{(10 \times {10^{ – 2}})^2} = 2.5\,J$
Standard 11
Physics
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