Work done by friction on an inclined plane

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Homework Help Overview

The discussion revolves around calculating the work done by friction on a crate being pushed up an inclined plane. The problem involves forces acting on the crate, including the worker's applied force, gravitational force, and frictional force, with specific values provided for weight and coefficients.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between the normal force and the forces acting on the crate, questioning how to calculate the normal force and its implications for work done by friction. There are inquiries about the distances traveled in relation to the forces involved.

Discussion Status

Participants are actively engaging with the problem, with some providing clarifications on the components of forces acting on the crate. There is a recognition of the need to resolve forces into components to understand the normal force better. Guidance has been offered regarding the equilibrium of forces.

Contextual Notes

The inclined plane is described as forming a right triangle with specific dimensions, which may influence calculations related to distances and forces. There is an acknowledgment of missing information regarding the normal force calculation, which is central to determining the work done by friction.

imatreyu
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Homework Statement



A worker pushes a crate weighing 93 N up an inclined plane. The worker pushes the crate horizontally, parallel to the ground.
a. The worker exerts a force of 85 N, how much work does he do? (A: 340 J)
b. How much work is done by gravity? (A: -280 J)
c. The coefficient of friction is 0.20. How much work is done by friction?



Homework Equations



W=Fd


The Attempt at a Solution


W=Fd
W= (mu*Fnormal)d


I'm completely stuck, as I don't know how to find the normal force!

Thank you in advance!
 
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I presume there is a diagram with additional data?
 
Oh! Sorry, well the inclined plane forms a right triangle: 3 meters tall, 4 meters long, 5 meter hypotenuse. .

But that's the only additional data.
 
imatreyu said:
Oh! Sorry, well the inclined plane forms a right triangle: 3 meters tall, 4 meters long, 5 meter hypotenuse. .

But that's the only additional data.
Thats all the extra data you need!
Part a) Whats the distance traveled by the crate along the direction of the worker's 85N force?
Part b) Whats the distance traveled by the crate along the direction of gravity?
Part c) Whats the distance traveled by the crate along the direction of friction?
All these numbers come from that very important inclined plane!
 
I'm confused?
How will the distance help in finding the amount of work done by friction?

Apparently, it's supposed to be done as such:

W= (mu*Fn)*d
W= mu (Fyou + Fg)d


I guess I should have been more clear. Specifically, I don't understand how (Fyou +Fg) equates to Fn. (Sorry about the confusion!)
 
The force that the worker exerts on the crate can be resolved into two components: along the slope and perpendicular to the slope. Similarly, the gravitational force on the crate can be resolved into two such components as well.
Analysing the crate in the direction perpendicular to the slope, we thus see that for the crate to remain in equilibrium in that direction, the normal contact force on the crate must balance out the components of the worker's force and gravity in that direction).
 
Ahh~!
Thank you so much! !
That made it really clear-- I get it now!
 

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