Question

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

Answer 1

`"distance" = 0.676` `"m "` (if `theta = pi/4`; see below)

Explanation:

I'd like to point out that there can't realistically be an inclined ramp with angle of inclination `(3pi)/4`...it should be between `0` and `pi/2`, so I'll choose the corresponding first quadrant angle to this, `pi/4`...

We're asked to find the distance traveled by the box given its initial speed, coefficient of kinetic friction, and angle of inclination.

NOTE: the friction force actually acts down the incline as opposed to the image (because the box is traveling up the ramp).

The magnitude of the kinetic friction force `f_k` is given by

`f_k = mu_kn`

where

• `mu_k` is the coefficient of kinetic friction (`5/3`)

• `n` is the magnitude of the normal force exerted by the incline plane, equal to `mgcostheta`

We must first find the acceleration of the box, using Newton's second law:

`sumF = ma`

`a = (sumF)/m`

The net horizontal force `sumF` is

`sumF = mgsintheta + f_k = mgsintheta + mu_kmgcostheta`

Therefore, we have

`a = (cancel(m)gsintheta + cancel(m)u_kmgcostheta)/(cancel(m)) = gsintheta + mu_kgcostheta`

Plugging in known values, we have

`a = (9.81color(white)(l)"m/s"^2)sin((pi)/4) + 5/3(9.81color(white)(l)"m/s"^2)cos((pi)/4)`

`= 18.5` `"m/s"^2`

directed down the incline, so this can also be written as

`a = ul(-18.5color(white)(l)"m/s"^2`

Now, we can use the equation

`(v_x)^2 = (v_(0x))^2 + 2a_x(Deltax)`

to find the distance it travels up the ramp before it comes to a stop.

Here,

• `v_x = 0` (instantaneously at rest at maximum height)

• `v_(0x) = 5` `"m/s"` (given initial velocity)

• `a_x = -18.5` `"m/s"^2`

• `Deltax =` trying to find

Plugging in known values, we have

`0 = (5color(white)(l)"m/s")^2 + 2(-18.5color(white)(l)"m/s"^2)(Deltax)`

`Deltax = color(red)(ul(0.676color(white)(l)"m"`