Potential Energy EX 8

The diagram below shows how the mass is being dropped onto the spring (fig. 1). It compresses the spring by 0.2 m (fig. 2) and then the spring decompresses while the mass moves up 0.1 m (fig. 3).

The compression of the spring changes from the initial compression of $x_{i}=0.2$ m to the final compression of the spring $x_{f}=0.1$ m. We expect to find the total change in potential energy to be negative when the mass is released because it moves spontaneously in this direction. The change in gravitational potential energy is positive since the mass moves up:

$\Delta U=mgh=(0.1)(10)(0.1)=0.1J$

The change in the spring's potential energy is

$\Delta U={\frac {1}{2}}k{x_{f}}^{2}-{\frac {1}{2}}k{x_{i}}^{2}={\frac {1}{2}}(50)(0.1)^{2}-{\frac {1}{2}}(50)(0.2)^{2}=-0.75J$

The total change in potential energy of this system is

$\Delta U=-0.75+0.1=-0.65J$