Question

The length of a wire increases by 1.0 cm when a body of mass 10.0 kg is suspended from the second end of the wire while the first end is clamped. Find the elastic potential energy stored in the wire
(g = 10 m/s²)


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Answer 1

kx = mg

k(0.1) = 10 *10

k=1000n/m

elastic potential energy is

U=1/2 k x^2

U= 0.5 *1000*0.1*0.1

U=500*1/10*10

U=5j