Work done by the Friction Force

[nobodyuknow]

Homework Statement

Find the work done by friction force as it slides down an inclined plane in terms of...

m (Mass)
x (Displacement)
μ (Friction Coefficient)
θ (Angle of the inclined plane)

Homework Equations

ƩFnormal = mgcos(θ)
ƩFhorizontal = mgsin(θ)
ƩFfriction = μFnormal - But then I realized these three equations all linked together which used gravity... which isn't suppose to be used)

F = ma
w = max

The Attempt at a Solution

What I came up with was:

Work = mgx(μcos(θ) + sin(θ))
Which doesn't seem right and uses gravity...

Help Please! Thanks.

 

[Doc Al]

g is just a constant, so why can't you use it? What's the basic definition of work done by a force?

 

[nobodyuknow]

Well, I'm not 100% sure, but it's basically, the amount of energy done by something?

In anycase, is my Attempt correct?

My friend did it, and he got Work = mgx(μcos(θ) - sin(θ))
not Work = mgx(μcos(θ) + sin(θ))

Cheers

 

[azizlwl]

http://en.wikipedia.org/wiki/Work_(physics)

1.In physics, mechanical work is a scalar quantity that can be described as the product of a force times the distance through which it acts, and it is called the work of the force

2.Joule (J) is the SI unit for work (defined as the product N*m, so that a joule is a Newton-metre).

 

[Doc Al]

Well, I'm not 100% sure, but it's basically, the amount of energy done by something?

I was looking for the simplest definition, such as Work = Force * displacement.

In anycase, is my Attempt correct?

My friend did it, and he got Work = mgx(μcos(θ) - sin(θ))
not Work = mgx(μcos(θ) + sin(θ))

Almost. Why do you include the sin(θ) term? Seems like you are trying to find the net work done, not just the work done by friction. (What direction does the friction force act compared to the displacement?)

What's the work done by the friction force alone?

 

[nobodyuknow]

Yes, you are correct, I am trying to find the net work done. Since friction works in the opposite direction of which work does, I assume my friend is correct with his formula.

Work = mgx(μcos(θ) - sin(θ))

Work = Fx
Friction Force = -Fx

 

[Doc Al]

Yes, you are correct, I am trying to find the net work done. Since friction works in the opposite direction of which work does, I assume my friend is correct with his formula.

Work = mgx(μcos(θ) - sin(θ))

If you want the net work done (by all forces) then that solution is almost correct. The sign is backwards: Since friction acts opposite to the displacement, its contribution to the net work will be negative.