A box with an initial speed of 8 m/s8ms is moving up a ramp. The ramp has a kinetic friction coefficient of 8/7 87 and an incline of ( pi )/4 π4. How far along the ramp will the box go?

1 Answer
Jun 3, 2017

The distance is =2.15m=2.15m

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*(8/7cos(1/4pi)+sin(1/4pi))a=9.8(87cos(14π)+sin(14π))

=-14.85ms^-2=14.85ms2

The negative sign indicates a deceleration

We apply the equation of motion

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

u=8ms^-1u=8ms1

v=0v=0

a=-14.85ms^-2a=14.85ms2

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

=(0-64)/(-2*14.85)=064214.85

=2.15m=2.15m