Calculating Work and Distance on an Inclined Plane

AI Thread Summary
A box with a mass of 4.0 kg is pushed up a 30-degree inclined plane using 128J of energy, while experiencing a kinetic friction of 0.30. The normal force is calculated to be 33.98N, leading to a frictional force of 10.194N. The gravitational force component acting down the incline is 19.6N, resulting in a net retarding force of 29.794N when both forces are combined. Using the work-energy principle, the distance traveled by the box until it stops is determined to be 4.29 meters. This calculation highlights the effects of friction and incline on the motion of the box.
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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 :(
 
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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
 
rl.bhat said:
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
 

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