A box with an initial speed of 3 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 5/7 and an incline of ( 5 pi )/12 . How far along the ramp will the box go?

1 Answer
Aug 14, 2017

"distance" = 0.399 "m"

Explanation:

We're asked to find how far the box will travel up the ramp, given its initial speed, the coefficient of kinetic friction, and the angle of inclination.

I will solve this problem using only Newton's laws and kinematics (i.e. without using work/energy).

I will take the positive x-direction as *up the ramp.*

When the box reaches its maximum distance, the instantaneous velocity will be 0. We are ultimately going to use the constant-acceleration equation

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

where

  • v_x is the instantaneous velocity (which is 0)

  • v_(0x) is the initial velocity

  • a_x is the (constant) acceleration

  • Deltax is the distance it travels (what we're trying to find)

Since the velocity v_x = 0, we can also write this equation as

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

We also figure that the acceleration will be negative because it slows down and comes to a brief stop at its maximum height, so we then have

0 = (v_(0x))^2 + 2(-a_x)(Deltax)

And we can move it to the other side:

ul(2a_x(Deltax) = (v_(0x))^2

Rearranging for the distance traveled Deltax:

ulbar(|stackrel(" ")(" "Deltax = ((v_(0x))^2)/(2a_x)" ")|)

We already know the initial velocity, so we need to find the acceleration of the box.

" "

Let's use Newton's second law of motion to find the acceleration, which is

ul(sumF_x = ma_x

where

  • sumF_x is the net force acting on the box

  • m is the mass of the box

  • a_x is the acceleration of the box (what we're trying to find)

The only forces acting on the box are

  • the gravitational force (acting down the ramp), equal to mgsintheta

  • the friction force (acting down the ramp), equal to mu_kn

And so we have our net force equation:

sumF_x = mgsintheta + mu_kn

The normal force n exerted by the incline is equal to

n = color(purple)(mgcostheta)

So we can plug this in to the net force equation above:

ul(sumF_x = mgsintheta + mu_kcolor(purple)(mgcostheta)

Or

ul(sumF_x = mg(sintheta + mu_kcostheta)

Now, we can plug this in for sumF_x in the Newton's second law equation:

sumF_x = ma_x

ma_x = mg(sintheta + mu_kcostheta)

We can cancel the mass m by dividing both sides by m, leaving us with

color(green)(ul(a_x = g(sintheta + mu_kcostheta)

" "

Now that we have found an expression for the acceleration, let's plug it into the equation

Deltax = ((v_(0x))^2)/(2a_x)

And we get

color(red)(ulbar(|stackrel(" ")(" "Deltax = ((v_(0x))^2)/(2g(sintheta + mu_kcostheta))" ")|)

We're given in the problem

  • v_(0x) = 3 "m/s"

  • theta = (5pi)/12

  • mu_k = 5/7

  • and the gravitational acceleration g = 9.81 "m/s"

Plugging these in:

color(blue)(Deltax) = ((3color(white)(l)"m/s")^2)/(2(9.81color(white)(l)"m/s"^2)(sin[(5pi)/12] + 5/7cos[(pi)/12])) = color(blue)(ulbar(|stackrel(" ")(" "0.399color(white)(l)"m"" ")|)