Question

A box with an initial speed of `8 m/s` is moving up a ramp. The ramp has a kinetic friction coefficient of `3/5` and an incline of `pi /4`. How far along the ramp will the box go?

Answer 1

Consider this scenario,

Moreover, assess the coordinate system as to be parallel with the incline. We are given this data,

`mu_k = 0.60`
`theta = pi/4 = 45°`
`v_0 = (8m)/s`

We need three vectors, and resolve gravity into its components due to the preceding coordinate system. A vertical vector pointing down, (i) gravity, a vector pointing up perpendicular to the incline, (ii) normal force, and a vector pointing to the left, (iii) `F_k` (friction).

Now, using Newton's second law, we resolve the gravity vector into its x and y components. (something like this, but friction is pointing left is well)

`SigmaF_y = F_N - mgcostheta = 0`
`therefore F_N = mgcostheta`

Recall,

`F_k = mu_k * F_N`

`SigmaF_x = -mu_k * mgcostheta - mgsintheta = ma`
`=> -mu_k*gcostheta - gsintheta = a`

Hence,

`a = -0.60 * (9.8m)/s^2 * cos(45°) - (9.8m)/s^2sin(45°) approx (-11.09m)/s^2`

Furthermore, recall,

`nu^2 = nu_0^2 + 2aDeltax`,

and assume the `nu = 0`, since when the box reaches the top it will stop and begin to slide back down as a result of the acceleration via gravity.

Hence,

`0 = (8.0m)/s - 2*(11.09m)/s^2*Deltax`
`therefore Deltax approx 0.36"m"`

Not very far, the kinetic friction is very high!