Question

A box with an initial speed of `9 m/s` is moving up a ramp. The ramp has a kinetic friction coefficient of `5/6` and an incline of `(3 pi )/8`. How far along the ramp will the box go?

Answer 1

The distance is `=3.33m`

Explanation:

Resolving in 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 of Motion

`F=m*a`

Where `a` is the acceleration of the box

So

`ma=-mu_kgcostheta-mgsintheta`

`a=-g(mu_kcostheta+sintheta)`

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

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

The incline of the ramp is `theta=3/8pi`

The acceleration is `a=-9.8*(5/6cos(3/8pi)+sin(3/8pi))`

`=-12.18ms^-2`

The negative sign indicates a deceleration

Apply the equation of motion

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

The initial velocity is `u=9ms^-1`

The final velocity is `v=0`

The acceleration is `a=-6.32ms^-2`

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

`=(0-81)/(-2*12.18)`

`=3.33m`