Potential Energy EX 4

The block slides down the distance 3 m and then continues to slide down some more. Let's assume that it slides some further distance $x$. It means that the total distance the block moves down the incline is $3+x$. It moves down the height $h=(3+x)\sin (30^{\circ })$. The spring which was initially uncompressed is now compressed by $x$.

The question is to find the distance $x$ such that the total change in potential energy of the system (spring-block-Earth) is zero. To find this distance we will first compute the change of height from the triangle in the diagram above:

$\sin (30^{\circ })=\frac{h}{3+x}$

$h=(3+x)\sin (30^{\circ })$

And then the change in gravitational potential energy using the equation:

$\mathrm{\Delta }{U}_G=-mgh=-(0.1)(10)((3+x)\sin (30^{\circ }))$

And then compute the change in the spring's potential energy:

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

The total change in potential energy is then

$\mathrm{\Delta }{U}_G+\mathrm{\Delta }{U}_{sp}=0$

$[-(0.1)(10)((3+x)\sin (30^{\circ }))]+[\frac{1}{2}(4)x^2]=0$

$-(3+x)+4x^2=0$

Solving the quadratic equation for $x$, we find $x=1$ m. Note that we select the positive solution since the distance is always positive.