Question

A stationary mass m = 1.3 kg is hanging from a spring of spring constant k = 1200 n/m. you raise the mass a distance of 10 cm above its equilibrium position. how much has the potential energy of the mass-spring system changed?

Answer 1

The weight of the mass is
$F=mg=(1.3 kg)(9.81 m/s^2)=12.8 N$
This force stretches the spring of a certain amount $x_1$, which can be calculated by using Hook's law:
$x_1 = \frac{F}{k} = \frac{12.8 N}{1200 N/m} =0.011 m$
with respect to the equilibrium position of the spring.
So, the initial elastic potential energy of the spring is
$U_1 = \frac{1}{2} kx_1^2 = \frac{1}{2} (1200 N/m)(0.011 m)^2=0.07 J$

Later, the spring is compressed by $x_2 = -10 cm =-0.100 m$ from its equilibrium position (the negative sign means that now it is a compression, while before it was an elongation). This means that the new elastic potential energy of the spring is
$U_2 = \frac{1}{2} kx_2^2 = \frac{1}{2}(1200 N/m)(-0.100 m)^2 = 6 J$

So, the change in elastic potential energy of the mass-spring system is
$\Delta U=U_2 -U_1 = 6J-0.07 J=5.93 J$

However, if we consider also the change in gravitational potential energy of the mass, the answer is different. In fact, the height of the mass changed by 
$\Delta h=x_1 -x_2 = 0.011 m-(-0.100 m)=0.111 m$
So its gravitational potential energy increased by
$\Delta E_{gp} = mg \Delta h=(1.3 kg)(9.81 m/s^2)(0.111 m)=1.42 J$

So, the total change of potential energy (elastic+gravitational) of the mass-spring system is
$\Delta U=5.93 J+1.42 J=7.35 J$