Question

4.82 kJ work is done by a worker pulling a crate across the deck of a ship at a constant speed of 7.82 m/s by a 453 N force. The rope is being pulled at a 37.0 degree angle to the deck.

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)Ffk=uk(Fn)
uk=Ffk/mg
uk=363N/264.6
uk=1.37

That's the best I can come up with but the coefficient of kinetic friction is awfully high and I'm also not sure that I calculated properly with regards to the normal force. Does that decrease due to the presence of the y component of the F applied? Thank you very much in advance!

Answer 1

a. W = F*d = 453*Cos37 * d = 4820 J.
d = 13.3 m.

b. Fx-Fk = M*a
453*Cos37-Fk = M*0 = 0
Fk = 453*Cos37 = 361.8 N. = Force of kinetic friction.

c. W = Fk*d = 361.8 * 13.3 = 4812 J.

d. W = M*V^2/2 = 4820 J.
M*7.82^2/2 = 4820
M*30.58 = 4820
M = 157.6 kg = Mass of the crate.

M*g = 157.6 * 9.8 = 1545 N. = Wt. of the
crate. = Normal force(Fn).

u = Fk/Fn = 361.8/1545 = 0.234

Answer 2

Thank you Henry! You did a really nice job and it helped me a lot!! (: