# Simple Work problem

## Homework Statement

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?

## Homework Equations

W = F$$_{par}$$d

## 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?

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

Delphi51
Homework Helper
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.

vela
Staff Emeritus