Work on 1.5x10^3 N Crate: Force, Displacement & Friction

[Alethia]

A worker pushes a 1.50 x 10^3 N crate with a horizontal force of 345 N a distance of 24.0 m. Assume the coefficient of kinetic friction between the crate and the floor is 0.220.

a) How much work is done by the worker on the crate?
b) How much work is fone by the floor on the crate?
c) What is the net work done on the crate?

--Okay, I understand that work equals force times the displacement times the cosin of theta. Howver, this "1.50 x 10^3 N crate" part is throwing me off! For part a, would I just use the formula and plug in 345 N or... see, I'm confused... someone care to explain? Thnks!

 

[jamesrc]

They give you the weight of the crate so you can find the friction force. Remember that there is a friction force opposing the motion and equal to μ_k*N, where N is the normal force (the reaction force of the floor to the weight of the crate).

 

[Alethia]

Ohhh, that makes sense. But why do I need the normal force?

For part a, I can solve it by using W=Fd, right?
So therefore it would be (345 N)(24.0 m) = 8280 Joules, right??

But then for part b, do I use the normal force as the force to be multiplied with the displacement?

Then for part c, after finding a and b, I can now calculate the net work done on the crate because part a is the horizontal component, and b is the vertical component... right?

Sorry, I'm not quite sure I understand completely.

 

[jamesrc]

Yes, part a is correct.

For part b, you use the same formula, but the force is the friction force. Friction is doing work against the worker over the length of the floor.

For the net work, you subtract part b from part a. The position of the crate in the vertical direction has not changed, so no work was done in that direction.

 

[HallsofIvy]

One of the problems you are having is apparent in your first post:
"I understand that work equals force times the displacement times the cosin of theta."

No, you don't understand that. You know the formula but you don't understand it! That is the formula for the work done in pushing weight up an inclined plane. The fact that there is no "theta" in this problem should have been a clue that that formula does not apply here.

Actually, I shouldn't be so harsh- the problem is confusing. It says "A worker pushes a 1.50 x 10^3 N crate with a horizontal force of 345 N a distance of 24.0 m. Assume the coefficient of kinetic friction between the crate and the floor is 0.220."

You are completely correct that the work done to move the crate is "force times distance"= 345*24= 8280 Joules since you are given the horizontal force.

However, to solve part "b) How much work is done by the floor on the crate?" you need to find the friction force which is 0.220*1500=
330 N. The floor does 330*24= 7920 Joules of work.

Of course, that means there was a "net" of 8280-7920= 360 Joules.

What has happened is that the person pushing the crate has used more force than necessary- 345- 330= 15 N more than necessary.
15*24= 360 Joules of work. Where did that work go? 7920 of work went into overcoming friction (and became heat). The remaining 360 Joules went into accelerating the box and became kinetic energy.

 

[Doc Al]

Excellent explanation, as usual, HallsofIvy. Let me add one minor point...

Originally posted by HallsofIvy
One of the problems you are having is apparent in your first post:
"I understand that work equals force times the displacement times the cosin of theta."

No, you don't understand that. You know the formula but you don't understand it! That is the formula for the work done in pushing weight up an inclined plane. The fact that there is no "theta" in this problem should have been a clue that that formula does not apply here.

I think part of the problem is that Work is often taught as W=Fd cosθ, but the meaning of θ is often forgotten as the formula is memorized. As long as θ is recognized as the angle between the displacement vector and the force vector, the "formula" works.

 

[Alethia]

Originally posted by HallsofIvy
One of the problems you are having is apparent in your first post:
"I understand that work equals force times the displacement times the cosin of theta."

No, you don't understand that. You know the formula but you don't understand it! That is the formula for the work done in pushing weight up an inclined plane. The fact that there is no "theta" in this problem should have been a clue that that formula does not apply here.

Actually, my teacher taught us to use that formula in all cases so, even if there is no theta. If there isn't she just told us to use the degree of 0 which equals one, therefore it doesn't change the problem. I guess it just depends...

Anyhow, thanks for all the help! It was very useful.