Question

A box with an initial speed of `9 m/s` is moving up a ramp. The ramp has a kinetic friction coefficient of `5/4` and an incline of `(3 pi )/4`. How far along the ramp will the box go?

Answer 1

The box will go `-23.36 m`etres along the ramp.

Explanation:

If a block is moving up the inclined plane then the distance travelled by the block up the plane is
`s={[u^2]/[2g(sintheta+mu_kcostheta)]}`
Assume box as block and ramp as inclined plane and also consider `g=9.8m//s^2`.
Calculating with given values,
`s={[9m//s]^2/[2xx(9.8m//s^2)xx[sin((3pi)/4)+(5/4)cos((3pi)/4)]]}`
`:.s=-23.36m`.
Here'`-`' sign denotes that the body is travelling up the incline plane.