Question

A box with an initial speed of `5 m/s` is moving up a ramp. The ramp has a kinetic friction coefficient of `1/2` and an incline of `(3 pi )/8`. How far along the ramp will the box go?

Answer 1

The distance is `=1.14m`

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`

`=-1/2*g*cos(3/8pi)-g*sin(3/8pi)`

`=-10.93ms^-2`

The initial velocity is `u=5ms^-1`

We apply the equation of motion

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

`0=5^2-2*10.93s`

`s=25/(2*10.93)`

`=1.14m`