A spring of spring constant 5 × 10^3 N/m is stretched initially by 5,cm from unstretched pos

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5.Work, Energy, Power and Collision

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A spring of spring constant $ 5 \times 10^3$ $ N/m$ is stretched initially by $5\,cm$ from the unstretched position. Then the work required to stretch it further by another $5\,cm$ is .............. $\mathrm{N-m}$

A

$6.25 $

B

$12.50$

C

$18.75$

D

$25$

(AIEEE-2003)

Solution

(c)$W = \frac{1}{2}k(x_2^2 – x_1^2) = \frac{1}{2} \times 5 \times {10^3}({10^2} – {5^2}) \times {10^{ – 4}}$
$ = 18.75\,J$

Standard 11

Physics

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