How High Does the Crate Reach on the Inclined Plane?

AI Thread Summary
The discussion focuses on calculating the maximum height reached by a crate sliding down a frictionless lunar crater and then up an inclined plane with kinetic friction. The key equation presented is mgy = mgh - (Mu sub k * n)y/sin(theta), but participants emphasize the necessity of determining the normal force (N) to solve the problem accurately. It is noted that there is zero acceleration normal to the incline, which is a crucial aspect of the analysis. The conversation highlights the importance of energy methods in solving the problem while addressing the nuances of the forces involved. Understanding the dynamics of the crate's motion is essential for finding the correct maximum height.
Leesh09
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Homework Statement



1. 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 as in Figure 2. 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





The Attempt at a Solution



The equation I have come up with so far is mgy=mgh-(Mu sub k *n)y/sin(theta)

I know this is correct, but how would one calculate what the normal force would be? or is it sufficient to simply solve for y?
 
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Hi Leesh09! :smile:

(btw, you shouldn't really call it g should you? :wink:)
Leesh09 said:
… how would one calculate what the normal force would be? or is it sufficient to simply solve for y?

Yes, you do need to know N.

Hint: there's zero acceleration normal to the incline. :wink:
 

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