Question

A box with an initial speed of `6 m/s` is moving up a ramp. The ramp has a kinetic friction coefficient of `1/2` and an incline of `(3 pi )/8`. How far along the ramp will the box go?

Answer 1

I found `1.6m` (but double check my maths!)

Explanation:

Consider the diagram:

We can use Newton's Second Law: `SigmavecF=mveca`
Where:
along the horizontal (along the ramp):
`-mgsin(3pi/8)-f=ma`
along the vertical:
`N-mgcos(5pi/8)=0`
`f=muN` is friction
`mgsin(5pi/8)` and `mgcos(5pi/8)` are the horizontal and vertical components of weight.

We also need to change `a` into an expression involving the distance `s`;
we can use kinematics and:
`v_f^2=v_i^2+2as`
`a=(v_f^2-v_i^2)/(2s)`
with
`v_i=6m/s`
`v_f=0`

So we get:
`-mgsin(3pi/8)-f=ma`
`-mgsin(3pi/8)-muN=m(v_f^2-v_i^2)/(2s)`
`-cancel(m)gsin(3pi/8)-cancel(m)mugcos(5pi/8)=cancel(m)(v_f^2-v_i^2)/(2s)`
`-9.8sin(3pi/8)-(1/2)9.8cos(5pi/8)=(0^2-6^2)/(2s)`
`-9.1-1.8=-36/(2s)`
`s=1/2(36/10.9)=1.6m`