Work done by friction on an inclined plane


by imatreyu
Tags: force, friction, inclined plane, work
imatreyu
imatreyu is offline
#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
Fightfish is offline
#2
Jan13-10, 03:30 AM
P: 595
I presume there is a diagram with additional data?
imatreyu
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#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
Fightfish is offline
#4
Jan13-10, 03:44 AM
P: 595

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
imatreyu is offline
#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
Fightfish is offline
#6
Jan13-10, 04:05 AM
P: 595
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
imatreyu is offline
#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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