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Work done by friction on an inclined plane

by imatreyu
Tags: force, friction, inclined plane, work
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imatreyu
#1
Jan13-10, 02:39 AM
P: 82
1. The problem statement, all variables and given/known data

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?



2. Relevant equations

W=Fd


3. 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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Fightfish
#2
Jan13-10, 03:30 AM
P: 623
I presume there is a diagram with additional data?
imatreyu
#3
Jan13-10, 03:33 AM
P: 82
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.

Fightfish
#4
Jan13-10, 03:44 AM
P: 623
Work done by friction on an inclined plane

Quote Quote by imatreyu View Post
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 travelled by the crate along the direction of the worker's 85N force?
Part b) Whats the distance travelled by the crate along the direction of gravity?
Part c) Whats the distance travelled by the crate along the direction of friction?
All these numbers come from that very important inclined plane!
imatreyu
#5
Jan13-10, 03:52 AM
P: 82
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!)
Fightfish
#6
Jan13-10, 04:05 AM
P: 623
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).
imatreyu
#7
Jan13-10, 04:13 AM
P: 82
Ahh~!
Thank you so much! !
That made it really clear-- I get it now!


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