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Elastic Potential Energy

Two similar s...

Question

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

A

W_{P} > W_{Q}; W_{Q} > W_{P}

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B

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

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C

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

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D

W_{P} = W_{Q}; W_{P} = W_{Q}

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Solution

The correct option is A $W_{P} > W_{Q}; W_{Q} > W_{P}$
Case a) both the springs are stretched by same amount, energy stored springs in P and Q will be

$W_{P} = \frac{1}{2} K_{P} Δ x^{2}$ and $W_{Q} = \frac{1}{2} K_{Q} Δ x^{2}$ ...(i)

given $K_{P} > K_{Q} \Rightarrow W_{P} > W_{Q}$ ...(ii)

Case b) both the springs are stretched by same force, energy stored springs in P and Q will be

$W_{P} = \frac{1}{2} K_{P} Δ x_{P}^{2}$ and $W_{Q} = \frac{1}{2} K_{Q} Δ x_{Q}^{2}$ ...(iii)

given $K_{P} > K_{Q} \Rightarrow Δ x_{P} < Δ x_{Q}$ ...(iv)
and $F = K_{P} Δ x_{P} = K_{Q} Δ x_{Q}$ ...(v)

$W_{P} = \frac{1}{2} K_{P} Δ x_{P} Δ x_{P}$ and $W_{Q} = \frac{1}{2} K_{Q} Δ x_{Q} Δ x_{Q}$ ...(vi)

from equation (v) substituting $F = K_{P} Δ x_{P} = K_{Q} Δ x_{Q}$ in equation (vi)

$W_{P} = \frac{1}{2} F x_{P}$ and $W_{Q} = \frac{1}{2} F x_{Q}$

given $Δ x_{P} > Δ x_{Q} \Rightarrow W_{P} < W_{Q}$ ...(vii)
hence, correct answer is option A.

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