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= (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

`(1)/(2) mv^2=mgh`

`h= (v^2)/(2g)= ((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= (h)/(\sin 20^(\circ))= (7.3 m)/(\sin 20^(\circ))=21.3 m`