Answer to Question #102173 in Mechanics | Relativity for dfd

Question

A 6.8 kg block slides down a 8.0 m long ramp inclined at an angle of 30o above the horizontal. At the bottom of the plane, the block transitions to a flat, smooth table that is 3.4 m in length. There is a spring attached to the wall at the end of the flat portion of the track that has spring constant of 255.0 N/m. What is the speed of the block when it leaves the spring on its return trip back across the track?

Answer 1

Since there is no energy loss, the potential energy (E_p) at the inclined plane will equal to the energy of the block leaving the spring (E_s) ,

$E_p​=E_s \\ E_p=mgh\\ E_p=mglsin(\theta)$

where l=inclined length of the plane,

$\theta$ is the inclination of the plane

$\frac{1}{2}mv^2=mglsin(\theta)\\ v=\sqrt{2lgsin(\theta)}\\ v=8.85ms^{-1}$