Question

A box with an initial speed of `5 m/s` is moving up a ramp. The ramp has a kinetic friction coefficient of `3/4` and an incline of `(2 pi )/3`. How far along the ramp will the box go?

Answer 1

The distance is `=2.6m`

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=3/4`

The acceleration due to gravity is `g=9.8ms^-2`

The incline of the ramp is `theta=2/3pi`

`a=-9.8*(3/4cos(2/3pi)+sin(2/3pi))`

`=-4.81ms^-2`

The negative sign indicates a deceleration

We apply the equation of motion

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

`u=5ms^-1`

`v=0`

`a=-4.81ms^-2`

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

`=(0-25)/(-2*4.81)`

`=2.6m`