Calculating Work and Distance on an Inclined Plane

[kanta]

Homework Statement

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 traveled until it stops?

Homework Equations

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