Calculating Work Done by Friction?

AI Thread Summary
A 3-kg block is moved up a 37° incline with a constant horizontal force of 40 N, and the coefficient of kinetic friction is 0.1. The main issue discussed is the calculation of work done by friction, where the initial attempt yielded -4.7 J instead of the correct -9.51 J. The error arose from not accounting for the additional normal force contributed by the horizontal force. A free body diagram is suggested to clarify the forces acting on the block. The discussion concludes with the participant acknowledging their misunderstanding and expressing gratitude for the clarification.
BrainMan
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Homework Statement


A 3-kg block is moved up a 37° incline under the action of a constant horizontal force of 40 N. The coefficient of kinetic friction is .1, and the block is displaced 2 m up the incline. Calculate (a) the work done by the 40-N force, (b) the work done by gravity, (c) the work done by friction, and (d) the change in kinetic energy of the block.


Homework Equations


The Potential and Kinetic Energy equations


The Attempt at a Solution


I managed to get all the parts of this question right except part (c). What I did was take the weight, which I found to be 29.4, and do 29.4 cos 37° to find the normal force. I got 23.48 for the normal force. Then I multiplied that by .1 to get the force of friction. Then I took the force of friction and multiplied it by the displacement to find the work done by friction. I got -4.7 J and the answer was -9.51 J.
 
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BrainMan said:

Homework Statement


A 3-kg block is moved up a 37° incline under the action of a constant horizontal force of 40 N. The coefficient of kinetic friction is .1, and the block is displaced 2 m up the incline. Calculate (a) the work done by the 40-N force, (b) the work done by gravity, (c) the work done by friction, and (d) the change in kinetic energy of the block.


Homework Equations


The Potential and Kinetic Energy equations


The Attempt at a Solution


I managed to get all the parts of this question right except part (c). What I did was take the weight, which I found to be 29.4, and do 29.4 cos 37° to find the normal force. I got 23.48 for the normal force. Then I multiplied that by .1 to get the force of friction. Then I took the force of friction and multiplied it by the displacement to find the work done by friction. I got -4.7 J and the answer was -9.51 J.
Draw a free body diagram for the block.

The horizontal force (actually the reaction to it) contributes to the normal force.
 
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BrainMan said:
I managed to get all the parts of this question right except part (c). What I did was take the weight, which I found to be 29.4, and do 29.4 cos 37° to find the normal force. I got 23.48 for the normal force.
That would be fine except for that 40 N horizontal force. That force has a component normal to the surface, so it adds to the normal force on the block.

(SammyS beat me to it!)
 
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OK I get it now. Thanks!
 
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