First slide

Work done by Diff type of forces

Question

Two springs A and B (K_A = 2K_B) are stretched by same suspended weight, The ratio of work done in stretching is :

Easy

A

1:2
B

2:1
C

1:1
D

1:4

Solution

work done $= \frac{1}{2} {Kx}^{2} = \frac{(Kx)^{2}}{2 K} = \frac{(mg)^{2}}{2 K}$

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