A box with an initial speed of 8 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 3/2 and an incline of pi /4 . How far along the ramp will the box go?

1 Answer
Nov 29, 2017

The distance is =2.77m

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=3/2

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

The incline of the ramp is theta=1/4pi

a=-9.8*(3/2cos(1/4pi)+sin(1/4pi))

=-11.57ms^-2

The negative sign indicates a deceleration

We apply the equation of motion

v^2=u^2+2as

u=8ms^-1

v=0

a=-11.57ms^-2

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

=(0-64)/(-2*11.57)

=2.77m