A box with an initial speed of 8 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 7/5 and an incline of (5 pi )/12 . How far along the ramp will the box go?

1 Answer
Jan 11, 2018

The distance is =2.46m

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

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=7/5

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

The incline of the ramp is theta= 5/12pi

a=-9.8*(7/5cos(5/12pi)+sin(5/12pi))

=-13.02ms^-2

The negative sign indicates a deceleration

We apply the equation of motion

v^2=u^2+2as

u=8ms^-1

v=0

a=-13.02ms^-2

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

=(0-64)/(-2*13.02)

=2.46m