Conservation of Mechanical Energy — Question 2

Question 2

Spring launch onto an inclined guide

Neglect friction and air resistance, take $g=10$ m/s$^2$, and treat all stated guides as fixed. A 0.50 kg slider is launched from rest by an unattached horizontal spring with $k=200$ N/m, initially compressed 0.15 m. After losing spring contact it climbs a straight $30^\circ$ incline. Treat the joining region as negligibly short and lossless. The incline base is at launch height.

1. Find the speed at the incline base.

2. Find maximum vertical rise and distance along the incline.

3. Explain why the spring potential is zero during the final climb.

Question 2 - Solution

Spring launch onto an inclined guide

(a) Convert the initial spring energy.

The spring releases $$S=\tfrac12k x_0^2=\tfrac12(200)(.15)^2=2.25\ \mathrm J.$$ $$\tfrac12(.50)v^2=S\Rightarrow\boxed{v=3\ \mathrm{m/s}}.$$

(b) Set kinetic energy to zero at the turning point.

At maximum rise, $K=0$: $$\boxed{h=\frac{S}{mg}=\frac{2.25}{.50(10)}=.45\ \mathrm m},\qquad\boxed{s=\frac{h}{\sin30^\circ}=.90\ \mathrm m}.$$

(c) Respect the contact condition.

The unattached spring separates at its relaxed length: $$x_s=0\Rightarrow U_s=\tfrac12kx_s^2=0,\qquad K+mgh=S\quad\text{during the climb}.$$ It cannot pull on the departing slider, so no later stretching energy belongs in the balance.