A box with an initial speed of 7ms is moving up a ramp. The ramp has a kinetic friction coefficient of 13 and an incline of π6. How far along the ramp will the box go?

1 Answer
May 31, 2018

The distance is =3.17m

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=13

The acceleration due to gravity is g=9.8ms2

The incline of the ramp is θ=16π

The acceleration is a=9.8(13cos(16π)+sin(16π))

=7.73ms2

The negative sign indicates a deceleration

Apply the equation of motion

v2=u2+2as

The initial velocity is u=7ms1

The final velocity is v=0

The acceleration is a=7.73ms2

The distance is s=v2u22a

=04927.73

=3.17m