First slide

Work done by Diff type of forces

Question

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

Easy

A

12.50 N-m
B

18.75 N-m
C

25.00 N-m
D

6.25 N-m

Solution

Work done = $= \frac{1}{2} k (x_{2}^{2} - x_{1}^{2})$
$= \frac{1}{2} (5 \times 10^{3}) [{(10 \times 10^{- 2})}^{2} - {(5 \times 10^{- 2})}^{2}] = 18 .75 N - m$

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