Question

8. Work of 500 J is done in stretching a spring by 2 m from its equilibrium position. Calculate its spring constant ​

Answer 1

Answer:

actually I don't know the answer

Answer 2

Answer:

Potential energy in a stretched spring  U=21kx2

where,  x is the extension in the spring.

So, U2=21kl22 and  U1=21kl12

Work done  W=U2−U1

W=21kl22−21kl12=21k(l22−l12).

Explanation:

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