ugh more physics homework please?

Question

A 10.0 kg crate is pulled up a rough incline with an initial speed of 1.5m/s. The pulling force is 100.0N parallel to the incline, which makes an angle of 15.0 degrees with the horizontal. Assuming the coefficient of kinetic friction is 0.40 and the crate is pulled a distance of 7.5 m, find the following:

a. the work done by the Earth's gravity on the crate
b. the work done by the force of friction on the crate
c. the work done by the puller on the crate
d. the change in kinetic energy of the crate
e. the speed of the crate after it is pulled 7.5 m

The answers for each one are (in order)
a. -1.9 X 10^2 J
b. -2.8 X 10^2 J
c. 7.5 X 10^2 J
d. 2.8 X 10^2 J
e. 7.6 m/s

if someone could show how to get these answers i would greatly appreciate it (again lol)

Answer 1

I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:

a)
Work = mgd *d=change in height
W = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J

b)
W=(friction coef)(natural force)(distance traveled)
natural force = (cos15)(10kg)(9.8m/ss)
W= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J

c)
W=Force x distance
W=(100N)(7.5m)=750J

d)
(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)
(final KE) - (initial KE) = 750J - 280J - 190J = 280J

e)
(final KE) - (initial KE) = 280J
*solve for final v
v = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s