1.

The work done in stretching a spring of constant K by amount by delta​

Answer»

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).



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