Question

A box with an initial speed of `2 m/s` is moving up a ramp. The ramp has a kinetic friction coefficient of `5/4` and an incline of `pi /6`. How far along the ramp will the box go?

Answer 1

The distance is `=0.13m`

Explanation:

Solving in the direction of the plane `↗^+`

`mu_k=F_k/N`

`F_k=mu_kN`

`N=mgcostheta`

`F_k=mu_kmgcostheta`

The component of the weight is

`=mgsintheta`

Applying Newton' second Law

`-mu_kmgcostheta-mgsintheta=ma`

Therefore,

`a=-u_kgcostheta-gsintheta`

`=-5/4*g*cos(1/6pi)-g*sin(1/6pi)`

`=-15.51ms^-2`

The initial velocity is `u=2ms^-1`

We apply the equation of motion

`v^2=u^2+2as`

`0=2^2-2*15.51s`

`s=4/(2*15.51)`

`=0.13m`