A box with an initial speed of 5 m/s5ms is moving up a ramp. The ramp has a kinetic friction coefficient of 3/7 37 and an incline of (3 pi )/8 3π8. How far along the ramp will the box go?

1 Answer
Apr 23, 2017

The distance is =1.17m=1.17m

Explanation:

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Solving in the direction of the plane ↗^++

mu_k=F_k/Nμk=FkN

F_k=mu_kNFk=μkN

N=mgcosthetaN=mgcosθ

F_k=mu_kmgcosthetaFk=μkmgcosθ

The component of the weight is

=mgsintheta=mgsinθ

Applying Newton' second Law

-mu_kmgcostheta-mgsintheta=maμkmgcosθmgsinθ=ma

Therefore,

a=-u_kgcostheta-gsinthetaa=ukgcosθgsinθ

=-3/7*g*cos(3/8pi)-g*sin(3/8pi)=37gcos(38π)gsin(38π)

=-10.66ms^-2=10.66ms2

The initial velocity is u=5ms^-1u=5ms1

We apply the equation of motion

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

0=5^2-2*10.66s0=52210.66s

s=25/(2*10.66)s=25210.66

=1.17m=1.17m