help please, how do I do part e? the speed of the crate using kinetic energy?

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 20.0° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.95 m.
(a) How much work is done by gravity?
-259.53
(b) How much mechanical energy is lost due to friction?
285.22 J
(c) How much work is done by the 100 N force?
595 J
(d) What is the change in kinetic energy of the crate?
50.26 J
(e) What is the speed of the crate after being pulled 5.95 m?


I got the right answer for the others, but I cant figure out how to do E, any help?

Answer 1

(a) Work done by gravity=W
W=F d= mg sin(20) 5.95=
W=13.0 x 9.81 sin(20) 5.95
W=259.5 joules (confirmed)

(b) mechanical energy is lost due to friction Ef
Ef=f d= u mg cos(20) 5.95=
Ef= 0.400 x 13.0 x 9.81 x cos(20) x 5.95=
Ef= 285.2 Joules confirmed
(c) confirmed
(d) confirmed
(e) Change in kinetic energy is dKe
dKe=Ke2-Ke1=50.26=.5mv2^2 - .5m v1^2)
v2=sqrt(2 dKe/m +v1^2)=sqrt(2 x 50.26/13 + (1.5)^2)= v2=3.16m/sec

Answer 2

Work = 1/2mv^2 - 1/2mv^2 where m is mass, the first v is the initial velocity, and the second v is the velocity 5.95 meters up. you know mass and the initial velocity, slove for v two. the work is the total work done on the crate, which you should be able to find pretty easily.