First slide

Work done by Diff type of forces

Question

Two similar springs P and Q have spring constants $K_{p} and K_{Q}$ such that Kp > $K_{Q}$ . They are stretched first by the same amount (case a), then by the same force (case b). The work done by the springs $W_{P} and W_{Q}$ are related as, in case (a) and case (b), respectively.

Moderate

A

$W_{P} = W_{Q}; W_{P} > W_{Q}$
B

$W_{P} < W_{Q}; W_{P} = W_{Q}$
C

$W_{P} > W_{Q}; W_{Q} > W_{P}$
D

$W_{P} < W_{Q}; W_{Q} < W_{P}$

Solution

Case (a): Springs are stretched by the same amount
The work done by the springs W = $\frac{1}{2} {Kx}^{2}$
Hence W $\propto K$
As $K_{P} > K_{Q} it means W_{P} > W_{Q}$
Case (b): Springs are stretched by the same force.
F = $K_{p} X_{p} = K_{Q} X_{Q}$
The work done by the springs
$W = \frac{1}{2} {Kx}^{2} = \frac{1}{2} K {(\frac{F}{K})}^{2} = \frac{1}{2} \frac{F^{2}}{K}$
Hence $W \propto \frac{1}{K}$ it means $W_{Q} > W_{P}$

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