A small lead sphere of mass m is hung from a spring of spring constant k...

A small lead sphere of mass m is hung from a spring of spring constant k. The
gravitational potential energy of the system equals zero at the equilibrium position of
the spring before the weight is attached.The total mechanical energy of the system
when the mass is hanging at rest is:

a . –kx^2.
b. –0.5kx^2.
c. 0.
d. 0.5kx^2
e. +kx^2.

Let's solve the problem.
When the shpere is hung, the potential enegry of the spring is W1 = kx^2/2,
gravitational potential energy is W2 = -mgx = -kx*x = - kx^2.
So, total mechanical energy of the system is
W = W1 + W2 = kx^2/2 - kx^2 = -0.5 kx^2
The answer is b.
Good luck!