Question

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?

Answer 1

`"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"" ")|)`