Question

4830J of work is done by a laborer pulling a 27.0 kg crate across a cement floor at a constant speed of 7.82 m/s by a 454 N force. The rope is being pulled at a 37.0 degree angle to the floor.

a)How far was it pulled?
b)What is the friction force?
c)What is the work done against friction?
d)What is the coefficient of kinetic friction.

a)W=Fdcos0
4830J=454N(x)cos37
d=13.3m

b)Vconstant
Fx=0
Fappx=Ffk=454Ncos37
Ffk=363N

c)Wffk=Ffk(d)
=363N(13.3m)
=4828J

d)u=?

I have been going in circles with this question in particular for 3 days now. Assuming I did parts a-c correctly, how do you find the coefficient of friction? (: Thanks

Answer 1

Normal force on floor = m g - 454 sin 37

friction force = mu * normal force

but friction force = 363 N you said
so

363 = mu (27*9.8 - 454 sin 37)

Answer 2

Thank you Damon for answering, and so fast! Isn't kinetic friction just the x component of the force applied with constant velocity or am I wrong? The final coefficient of friction comes out to -42. which definitely is weird right? Could this problem be bogus? Thank you for your time!!

Answer 3

It only weighs about 270 N if 27 kg
He is pulling with 454 N

That is not right unless it is hooked to an anchor.
It should probably be 270 kg for about 2700 N weight

Answer 4

You were right! The weight was indeed wrong, the problem was incorrect! Thank you again!!