A box with an initial speed of 7ms is moving up a ramp. The ramp has a kinetic friction coefficient of 43 and an incline of π6. How far along the ramp will the box go?

1 Answer
Jan 14, 2018

The distance is =1.51m

Explanation:

Resolving in the direction up and parallel to the plane as positive +

The coefficient of kinetic friction is μk=FrN

Then the net force on the object is

F=FrWsinθ

=Frmgsinθ

=μkNmgsinθ

=mμkgcosθmgsinθ

According to Newton's Second Law

F=ma

Where a is the acceleration

So

ma=μkgcosθmgsinθ

a=g(μkcosθ+sinθ)

The coefficient of kinetic friction is μk=43

The acceleration due to gravity is g=9.8ms2

The incline of the ramp is θ=16π

a=9.8(43cos(16π)+sin(16π))

=16.22ms2

The negative sign indicates a deceleration

We apply the equation of motion

v2=u2+2as

u=7ms1

v=0

a=9.3ms2

s=v2u22a

=049216.22

=1.51m