Work done along incline plane

[kanta]

1. The problem statement, all variables and given/known data
There is a plane inclined 30degree. A box with 4.0kg mass is pushed up the plane with 128J. The kinetic friction between the plane and the box is 0.30. What is the distance will the box travelled until it stops?


2. Relevant equations



3. The attempt at a solution
Fn=MG Cos theta
= 4.0 x 9.8 x cos30
= 33.98N

Friction force = 33.98N x 0.30
= 10.194N

F = mg sin theta
= 4 x 9.8 x sin30
= 19.6N

19.6N - 10.194N = 9.406N

E = Fd

128J = 9.406N x d
d = 13.61M


i need help urgently :(

[kanta]

anyone? pls.........

[rl.bhat]

Mass is pushed up the plane. Therefore the frictional force acts in the direction of mg sin theta. Hence net retarding force is = 19.6N + 10.194N

[kanta]

Mass is pushed up the plane. Therefore the frictional force acts in the direction of mg sin theta. Hence net retarding force is = 19.6N + 10.194N

yeah, i got that , the displacement is 4.29meter

thx a lot