# Simple Work problem

1. Feb 28, 2010

### lockedup

1. The problem statement, all variables and given/known data
Suppose a worker pushes down at an angle of 30° below the horizontal on a 30.0‐kg crate
sliding with along a horizontal warehouse floor (μ$$_{k}$$ = 0.25) for 4.5 meters.
a. What magnitude of force must the worker apply to move the crate at constant velocity?

b. How much work is done on the crate by this force over the 4.5 m?

c. How much work is done on the crate by friction over that same distance?

d. How much work is done on the crate by the normal force over this distance?

e. How much work is done on the crate by gravity?

f. What is the total work done on the crate?

2. Relevant equations
W = F$$_{par}$$d

3. The attempt at a solution
(d), (e), and (f) are pretty simple. The normal force and gravity are perpendicular to motion so don't affect work. Since it's moving at a constant velocity, the net force is zero, so there's no work there either.

My problem is really the distinction between (b) and (c). I calculated the frictional force to 73.5 N. For part (a), the frictional force is just the horizontal component of the total force needed to push the crate. But only the frictional force will affect the amount of work done, right? So (b), and (c) are the same answer?

2. Feb 28, 2010

### shallgren

Remember: friction always opposes motion. The block is going along the floor, but friction is acting upon it trying to push it backwards.

3. Feb 28, 2010

### Delphi51

I agree that (b) and (c) are the same.
But I don't think you have the correct answer for (a).
The trouble is that the harder you push at the 30 degree angle, the harder your vertical component pushes the crate against the floor, and the more friction you get. That is, in F = μ*Fn, the detailed expression for Fn involves F. It isn't too bad to work out, but you do have to consider that vertical component of the applied force.

4. Feb 28, 2010

### vela

Staff Emeritus
Not exactly. What you wrote is a little bit confusing. The horizontal component of the applied force is equal in magnitude to the frictional force but points in the opposite direction. It is not the frictional force. I think that's what you meant by what you wrote.

You're right that it's only the horizontal component of the applied force which does work; the vertical component, like the normal force and gravity, is perpendicular to the direction of motion so it does no work.

Think about the sign of the work done by the applied force and the work done by friction.

5. Mar 1, 2010

### lockedup

Thank you, vela. I did mean to say that the horizontal component of the applied is equal in magnitude to the frictional force.