A box with an initial speed of 9 m/s9ms is moving up a ramp. The ramp has a kinetic friction coefficient of 1/2 12 and an incline of (2 pi )/3 2π3. How far along the ramp will the box go?

1 Answer
Jul 31, 2017

"distance" = 3.70distance=3.70 "m"m

Explanation:

I'd like to point out that there can't realistically be an inclined ramp with angle of inclination (2pi)/32π3...it must be between 00 and pi/2π2, so I'll choose the closest angle to this, pi/3π3...

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

upload.wikimedia.orgupload.wikimedia.org

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_kfk is given by

f_k = mu_knfk=μkn

where

  • mu_kμk is the coefficient of kinetic friction (1/212)

  • nn is the magnitude of the normal force exerted by the incline plane, equal to mgcosthetamgcosθ

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

sumF = maF=ma

a = (sumF)/ma=Fm

The net horizontal force sumFF is

sumF = mgsintheta + f_k = mgsintheta + mu_kmgcosthetaF=mgsinθ+fk=mgsinθ+μkmgcosθ

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)/3) + 1/2(9.81color(white)(l)"m/s"^2)cos((pi)/3)

= 10.9 "m/s"^2

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

a = ul(-10.9color(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) = 9 "m/s" (given initial velocity)

  • a_x = -10.9 "m/s"^2

  • Deltax = trying to find

Plugging in known values, we have

0 = (9color(white)(l)"m/s")^2 + 2(-10.9color(white)(l)"m/s"^2)(Deltax)

Deltax = color(red)(ul(3.70color(white)(l)"m"