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Work Energy and Power question

2015 · Q142
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Work Energy and Power question

2015 · Q142

NEETPhysicsWork Energy and PowerMCQ+4 / −1
Two similar springs P and Q have spring constants KP and KQ, such that KP > KQ. They are stretched first by the same amount (case a), then by the same force (case b). The work done by the springs WP and WQ are related as, in case (a) and case (b) respectively
  1. A
    WP > WQ;  WQ > WP
  2. B
    WP < WQ;  WQ < WP
  3. C
    WP = WQ;  WP > WQ
  4. D
    WP = WQ;  WP = WQ
View written solutionFree

Correct answer: A

Here, KP > KQ

Case (a) : Elongation (x) in each spring is same.

WP=12KPx2,WQ=12KQx2{W_P} = {1 \over 2}{K_P}{x^2},{W_Q} = {1 \over 2}{K_Q}{x^2}WP​=21​KP​x2,WQ​=21​KQ​x2
∴WP>WQ \therefore {W_P} \gt {W_Q}∴WP​>WQ​

Case (b) : Force of elongation is same.

So, x1=FKP{x_1} = {F \over {{K_P}}}x1​=KP​F​ and x2=FKQ{x_2} = {F \over {{K_Q}}}x2​=KQ​F​

WP=12KPx12=12F2KP{W_P} = {1 \over 2}{K_P}x_1^2 = {1 \over 2}{{{F^2}} \over {{K_P}}}WP​=21​KP​x12​=21​KP​F2​

WQ=12KQx22=12F2KQ{W_Q} = {1 \over 2}{K_Q}x_2^2 = {1 \over 2}{{{F^2}} \over {{K_Q}}}WQ​=21​KQ​x22​=21​KQ​F2​
∴WP<WQ \therefore {W_P} \lt {W_Q}∴WP​<WQ​

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