How High Does the Crate Reach on the Lunar Crater's Inclined Plane?

[Leesh09]

Homework Statement

A crate with scientific equipment slides down a curved frictionless side of lunar crater of the depth h, and then up along the other side which is an inclined plane. The coefficient of kinetic friction between crate and incline is k, and the inclined side makes an angle θ with the horizontal. Use energy methods to find the maximum height ymax reached by the crate.

Homework Equations

Ug=mgy
I know there are others but I'm not sure what is relevent

The Attempt at a Solution

Right now I have something like mgy=mgh-(k*n)y/sin theta.

I'm pretty sure this is terribly wrong but I need help with some explanation to it. Please!

 

[PhanthomJay]

That looks right, you just need to calculate the normal force, which you have designated as 'n'.

 

[tomcue]

First, let's start by defining the variables in the problem:

m = mass of the crate
g = acceleration due to gravity (on the moon)
y = height of the crate above the ground
h = depth of the lunar crater
k = coefficient of kinetic friction
n = normal force
θ = angle of the inclined plane with the horizontal

Now, we can use the conservation of energy principle to solve this problem. At the highest point, the crate will have potential energy and at the lowest point, it will have kinetic energy. We can set these two energies equal to each other to find the maximum height ymax reached by the crate:

Potential energy at highest point: mgy
Kinetic energy at lowest point: (1/2)mv^2

Since the crate is sliding down a frictionless surface, we can assume that all of its potential energy is converted into kinetic energy at the bottom of the crater. However, when the crate starts sliding up the inclined plane, it will experience friction, which will decrease its kinetic energy. We can account for this by subtracting the work done by friction from the kinetic energy:

(1/2)mv^2 - work done by friction = mgy

To find the work done by friction, we first need to calculate the normal force n. This can be done using Newton's second law:

n = mgcosθ

Now, we can substitute this value into our equation:

(1/2)mv^2 - (k*n)y = mgy

We can simplify this equation by dividing both sides by m and substituting the value of n:

(1/2)v^2 - ky(cosθ)y = gy

Next, we can use some trigonometry to simplify the equation further. We know that cosθ = h/y, so we can substitute this into the equation:

(1/2)v^2 - ky(h/y)y = gy

Finally, we can solve for the maximum height ymax by setting the derivative of this equation with respect to y equal to 0:

(1/2)(d/dy)v^2 - k(h/y) = g

Solving for y, we get:

y = √(2gh/(1+2kcosθ))

Therefore, the maximum height ymax reached by the crate is √(2gh/(1+2kcosθ)).