Question

A box with an initial speed of `5 m/s` is moving up a ramp. The ramp has a kinetic friction coefficient of `3/7` and an incline of `(5 pi )/8`. How far along the ramp will the box go?

Answer 1

`"distance" = 1.17` `"m"`

Explanation:

I'd like to point out that there can't realistically be an inclined ramp with angle of inclination `(5pi)/8`...it must be between `0` and `pi/2`, so I'll choose the closest angle to this, `(3pi)/8`...

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 (`3/7`)

• `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((3pi)/8) + 3/7(9.81color(white)(l)"m/s"^2)cos((3pi)/8)`

`= 10.7` `"m/s"^2`

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

`a = ul(-10.7color(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"`

• `a_x = -10.7` `"m/s"^2`

• `Deltax =` trying to find

Plugging in known values, we have

`0 = (5color(white)(l)"m/s")^2 + 2(-10.7color(white)(l)"m/s"^2)(Deltax)`

`Deltax = color(red)(ul(1.17color(white)(l)"m"`