Question

When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L + l). The elastic potential energy stored in the extended wire is:

(A) \(\frac{1}{2}\)Mgl

(B) \(\frac{1}{2}\)MgL

(C) Mgl

(D) MgL

Answer 1

(A) \(\frac{1}{2}\)Mgl

elastic potential energy: \(\frac{1}{2}\) × F × L = \(\frac{1}{2}\)Mgl