Block projected up inlcline with initial speed

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Homework Help Overview

The problem involves a block projected up an inclined plane at a 40-degree angle with an initial speed of 2 m/s and a coefficient of friction of 0.05. The objective is to determine the time it takes for the block to ascend and then return to its starting point.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to establish the forces acting on the block, including friction and gravitational components, and is working towards setting up the equations of motion. Some participants question the signs associated with the forces, particularly regarding the direction of friction relative to the block's motion.

Discussion Status

Participants have provided guidance on the correct application of forces and the importance of sign conventions. The discussion is ongoing, with the original poster refining their equations based on feedback.

Contextual Notes

The original poster is grappling with the setup of their equations and the implications of friction and gravitational forces, indicating a need for clarity on these concepts.

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Homework Statement


I have a block on an inclined plane (the angle is 40). The block is projected with an initial speed of 2m/s and μ=0.05. I need to find the time it takes the block to go up the inclined plane and return to the point it started out.

Homework Equations

The Attempt at a Solution


I know that the force of friction will be μmgcosθ. Usually, the force bringing the block down the plane is mgsinθ.
I figure that I will need to find the acceleration and then use the equations of motion to find t, since the final speed at the top of the plane will be 0. I am having trouble setting up my equations.
Would it be that:
ma = -mgsinθ+ μmgcosθ
 
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You're on the right track. Keep in mind that friction always opposes the direction of motion. So while the block is moving upslope, friction is directed downslope as is the relevant component of the gravitational force. In other words, be careful of the signs that you apply to the friction term.
 
gneill said:
You're on the right track. Keep in mind that friction always opposes the direction of motion. So while the block is moving upslope, friction is directed downslope as is the relevant component of the gravitational force. In other words, be careful of the signs that you apply to the friction term.
Ok, I've been working on it and I think that going up the plane it will be: ma = -mgsinθ-μmgcosθ and going down the plane it would be ma=-mgsinθ+μmgcosθ since I put the axis as being x directed up the plane and y directed like the normal
 
Looks good so far!
 

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