Force pushing 184-kg crate with net work zero by friction

[Miss1nik2]

A 184-kg crate is being pushed across a horizontal floor by a force P that makes an angle of 15.2 ° below the horizontal. The coefficient of kinetic friction is 0.264. What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?

Ok so this is what I've come up with...

w=mg= (184)(9.8) = 1803.2 N

Fk= Uk * Fn = (.264)(1803.2) = 476.05

Fn equals w (but opposite, right?

Then...

Fnet of y = 0

Fnet of x = Pcos(15.2) - Fk = 0

so...

P = 476.05 / cos(15.2) = 493.30 N.



But I am told that's the wrong answer! Why? Please help me!

Thank you!

 

[Redbelly98]

The normal force is not simply equal to the weight, since P has a downwards component.

Have you drawn a freebody diagram? That's the best way to see all forces involved.

 

[krausr79]

The push force is partially up and partially sideways. The up part of the force counteracts some of the gravity force and changes the normal force.