# How much work is done on the crate by friction?

A factory worker pushes a 35.0-kg crate a distance of 5.0 m along a level floor at constant speed by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25. (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by this force? (c) How much work is done on the crate by friction? (d) How much work is done my the normal force? By gravity?
Ok so because the crate is being pushed at constant speed, $$\Sigma F_{x} = 0$$ and $$\Sigma F_{y} = 0$$. For part (a) wouldnt the force the worker has to apply be equal to $$f_{k} = \mu_{k} n = (0.25)(343 N) = 85.75 N$$?

(b) Since $$W = f\bullet s$$, $$W = 85.75\times\cos 0\times(5.0 m) = 428.75 N$$. In part (c) would this just be the same answer, but opposite sign?

(d) The work would be 0, because the forces are perpendicular to the path of the box?

Thanks

Last edited:

daniel_i_l
Gold Member
Remember that forces at right angles to the direction of movement don't do any work.

If the worker pushes downward at an angle of 30 degrees below the horizontal, (a) What magnitude of force must the worker apply to move the crate at constant speed?

SinceI know the the horizontal component of the force is $$85.75 N$$, shouldn't we be able to find the force by: $$F\cos 30 = 85.75 N$$? When I solve for F I get about 100N. The answer is 116N. What am I doing wrong?

Thanks

daniel_i_l
Gold Member
How does the y-component of the force effect the friction?

The y-component in addition to the weight of the crate serve as the normal force. In other words, $$f_{k} = \mu_{k}(F_{y}+w)$$. Is this correct?

So would the force equal the friction forcE?

any ideas?

thanks

I just plugged in the correct answer, 116N and received $$116\times cos 30 = 100$$. I was getting 100 before.