A box with an initial speed of 6 m/s6ms is moving up a ramp. The ramp has a kinetic friction coefficient of 5/3 53 and an incline of (3 pi )/8 3π8. How far along the ramp will the box go?

1 Answer
May 4, 2017

The distance is =1.18m=1.18m

Explanation:

Taking the direction up and parallel to the plane as positive ↗^++

The coefficient of kinetic friction is mu_k=F_r/Nμk=FrN

Then the net force on the object is

F=-F_r-WsinthetaF=FrWsinθ

=-F_r-mgsintheta=Frmgsinθ

=-mu_kN-mgsintheta=μkNmgsinθ

=mmu_kgcostheta-mgsintheta=mμkgcosθmgsinθ

According to Newton's Second Law

F=m*aF=ma

Where aa is the acceleration

So

ma=-mu_kgcostheta-mgsinthetama=μkgcosθmgsinθ

a=-g(mu_kcostheta+sintheta)a=g(μkcosθ+sinθ)

a=-9.8*(5/3cos(3/8pi)+sin(3/8pi))a=9.8(53cos(38π)+sin(38π))

=-15.3ms^-2=15.3ms2

The negative sign indicates a deceleration

We apply the equation of motion

v^2=u^2+2asv2=u2+2as

u=3ms^-1u=3ms1

v=0v=0

a=-ms^-2a=ms2

s=(v^2-u^2)/(2a)s=v2u22a

=(0-36)/(-2*15.3)=036215.3

=1.18m=1.18m