Given, mass of block, m=1‌kg Spring constant, k=100N∕m In case I, extension in spring, Δx1=10‌cm =10×10−2m Initial speed, u=0‌ms−1 In case II extension in spring Δx2=5‌cm =5×10−2m Let, kinetic energy and potential energy be (KE) and (PE) As we know that, KE=
1
2
k(Δx12−Δx22) =
1
2
×100(102−52)×10−4 =50×75×10−4=0.375J and PE=