A box with a mass of 2 kg and initial speed of 3 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 2/3 and an incline of pi /12 . How far along the ramp will the box go?

1 Answer
Nov 29, 2017

The distance is =0.23m

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=2/3

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

The incline of the ramp is theta=1/12pi

a=-9.8*(2/3cos(1/12pi)+sin(1/12pi))

=-8.85ms^-2

The negative sign indicates a deceleration

We apply the equation of motion

v^2=u^2+2as

u=6ms^-1

v=0

a=-8.85ms^-2

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

=(0-4)/(-2*8.85)

=0.23m