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:

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

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:

$\mathrm{\Delta }U=\frac{1}{2}k{x_f}^2-\frac{1}{2}k{x_i}^2=\frac{1}{2}k{x}^2=10x^2$

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

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

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