A box with an initial speed of 4ms is moving up a ramp. The ramp has a kinetic friction coefficient of 16 and an incline of π3. How far along the ramp will the box go?

1 Answer
Jan 27, 2018

The distance is =0.73m

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 of Motion

F=ma

Where a is the acceleration of the box

So

ma=μkgcosθmgsinθ

a=g(μkcosθ+sinθ)

The coefficient of kinetic friction is μk=16

The acceleration due to gravity is g=9.8ms2

The incline of the ramp is θ=13π

a=9.8(16cos(13π)+sin(13π))

=10.89ms2

The negative sign indicates a deceleration

We apply the equation of motion

v2=u2+2as

u=4ms1

v=0

a=9.7ms2

s=v2u22a

=016210.89

=0.73m