Question

A block is given a very brief push up a 20.0° frictionless incline to give it an initial speed of 12.0 m/s. (a) how far along the surface of the plane does the block slide before coming to rest? m

Answer 1

The incline is frictionless, this means we can use the conservation of energy: the initial kinetic energy of the block
$K= \frac{1}{2}mv^2$
is converted into gravitational potential energy
$U=mgh$
where h is the height reached by the block as it stops. By equalizing the two formulas, we get
$\frac{1}{2} mv^2=mgh$
$h= \frac{v^2}{2g}= \frac{(12.0 m/s)^2}{2(9.81 m/s^2)} =7.3 m$

However, this is the maximum height reached by the block. The distance along the surface of the plane is given by:
$d= \frac{h}{\sin 20^{\circ}}= \frac{7.3 m}{\sin 20^{\circ}}=21.3 m$