What is the change in kinetic energy and the final velocity?

Question

A 13 kg crate is pulled by a force (parallel to the incline) up a rough incline. The crate has an initial speed of 1.69 m/s. The crate is pulled a distance of 7.41 m on the incline by a 150 N force. The coefficient of friction is 0.328. The acceleration of gravity is 9.8 m/s^2.

(Please help I haven't had any luck with this problem.)

Answer 1

Let's use energy equations to solve
the initial KE1 is
.5*13*1.69^2
18.56 J

The work done by the force is
150*7.41
1111.5 J

What you are missing is the angle of the incline
I will call it th

The gain in PE is
13*9.8*sin(th)*7.41
944*sin(th) J

The work done by friction is
13*9.8*cos(th)*0.328
41.79*cos(th) J

the change is KE is
KE2-KE1

KE2-18.56=
1111.5-944*sin(th)-41.79*cos(th)


the final velocity can be found from

.5*m*v^2=
1111.5-944*sin(th)-41.79*cos(th)+18.56
solve for v

j

Answer 2

finished capability till now result is TE = ke(A) = a million/2 mV^2; the place m = .25 kg and V = .798 mps previous to colliding with B. all of the capability is kinetic capability belonging to %. A. TE = ke(A) + ke(B) + we = a million/2 mv^2 + a million/2 Mv'2 + we; the place M = .350, v = .119 mps, and v' = .655 mps in the process the end result and we is the artwork capability (losses through deformation and friction) in the process the end result. we = ? is the exchange (loss) in kinetic capability you're in seek of. for this reason, a million/2 mV^2 - a million/2mv^2 - a million/2 Mv'^2 = we = (a million/2)*(.250*(.798^2 - .119^2) - .350*.655^2) = 0.002751Newton-m of kinetic capability lost to deformation and friction. that's a conservation of capability problem. the finished capability of %. A is converted to the two lesser KE's of A and B blended, plus the deformation and friction losses in the process the end result. Sum up the KE's after the end result and subtract that from the KE previous to result and the variation is the loss in kinetic capability you're in seek of.