A crate of mass 13.0 kg is pulled up a rough incline with an initial speed of 1.

Question

A crate of mass 13.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 19.5° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.45 m.

I have solved a,b and c. I need d and e

(a) How much work is done by gravity?
-231.77 J

(b) How much mechanical energy is lost due to friction?
261.80 J

(c) How much work is done by the 100 N force?
545 J

(d) What is the change in kinetic energy of the crate?

Your response differs from the correct answer by more than 100%. J
(e) What is the speed of the crate after being pulled 5.45 m?
m/s

Explanation / Answer

Net workdone by all forces acting on the crate is W = (Fnet)(S) = (545 J - 231.77 J - 261.80 J)(5.45 m) = 280.2935 J According to work-energy theorem W = KE Hence the change in the kinetic energy of the crate is W = KE = 280.2935 J = 280.3 J
(e) According to work-energy theorem W = KE = KEf - KEi = (1/2)mv2 - (1/2)mu2 280.3 J = (1/2)mv2 - (0.5)(13.0 kg)(1.50 m/s)2 (1/2)mv2 = 280.3 J + (0.5)(13.0 kg)(1.50 m/s)2 = 294.925 J v = 6.74 m/s   

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