Potential Energy EX 3

When released, the block of mass moves down and therefore the gravitational potential energy of the system (Earth-block-spring) decreases. Consequently, the change in gravitational potential energy is negative. While the block slides down some distance $x$, the spring stretches the same distance. Consequently, the potential energy of the spring increases. The change of the gravitational potential energy can be computed using:

$\Delta U=-mgh=-(3)(10)x=-30x$

The initial extension of the spring is $x_{i}=0$ and the final extension of the spring is $x_{f}=x$. We can compute the change in potential energy of the spring as follows:

$\Delta U={\frac {1}{2}}k{x_{f}}^{2}-{\frac {1}{2}}k{x_{i}}^{2}={\frac {1}{2}}kx^{2}=10x^{2}$

The total potential energy in this case is equal to zero and so

$\Delta U=-30x+10x^{2}=0$

Solving for $x$, we find $x=3$ m.