Question

A box with an initial speed of `3 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

The distance is `=0.36m`

Explanation:

Taking the direction up and parallel to the plane as positive `↗^+`

The coefficient of kinetic friction is `mu_k=F_r/N`

Then the net force on the object is

`F=-F_r-Wsintheta`

`=-F_r-mgsintheta`

`=-mu_kN-mgsintheta`

`=mmu_kgcostheta-mgsintheta`

According to Newton's Second Law

`F=m*a`

Where `a` is the acceleration
So

`ma=-mu_kgcostheta-mgsintheta`

`a=-g(mu_kcostheta+sintheta)`

The coefficient of kinetic friction is `mu_k=5/6`

The incline of the ramp is `theta=1/3pi`

`a=-9.8*(5/6cos(1/3pi)+sin(1/3pi))`

`=-12.57ms^-2`

The negative sign indicates a deceleration

We apply the equation of motion

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

`u=3ms^-1`

`v=0`

`a=-12.7ms^-2`

`s=(v^2-u^2)/(2a)`

`=(0-9)/(-2*12.57)`

`=0.36m`