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.

Question

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!!