Answer to Question #102171 in Mechanics | Relativity for bjbnl

Question

A 9.5 kg block slides down a 13.2 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 2.1 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 256.0 N/m. What is the velocity of the box when it reaches the bottom of the ramp?

Answer 1

$mgh=\frac{mv^2}{2}$

$mgl\sin 30°=\frac{mv^2}{2}$

$v^2=2gl\sin30°$

$v=\sqrt{2gl\sin30°}=\sqrt{2\cdot 9.81\cdot 13.2\cdot \sin30°}=11.38m/s$