Question

A box with an initial speed of `4 m/s` is moving up a ramp. The ramp has a kinetic friction coefficient of `5/6` and an incline of `pi /3`. How far along the ramp will the box go?

Answer 1

`s=v^2/(2g(sin alpha+ k cos alpha))`

`s=0.64m`

Explanation:

`E_k=E_p+W_T`
`mv^2/2=mgh+F_ts`
`F_t=F_n=mg*cos alpha`
`h/s=sin alpha`
`mv^2/2=mgs *sin alpha+kmgs cos alpha`
`s=v^2/(2g(sin alpha+ k cos alpha))`
`k=5/6`
`alpha=pi/3`
`g=9.81m/s^2`
`s=0.64m`