A factory worker pushes a 30.1 kg crate a distance of 4.1 m along a level floor at constant velocity by pushing downward at an angle of 28 ∘ below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.24.
1-What magnitude of force must the worker apply to move the crate at constant velocity?
2-How much work is done on the crate by this force when the crate is pushed a distance of 4.1m?
3-How much work is done on the crate by friction during this displacement?
4-How much work is done by the normal force?
5-How much work is done by gravity?
6-What is the total work done on the crate?
1] Let applied force be F. For constant velocity net horizontal force = 0
F*cos theta - friction = 0
F*cos 28 degree - u*(mg+F*sin 28 degree) = 0
F*cos 28 degree - 0.24*(30.1*9.8+F*sin 28 degree) = 0
F*cos 28 degree - 0.24*(30.1*9.8+F*sin 28 degree) = 0
0.882947*F -0.24*(30.1*9.8+F*0.46947) = 0
F = 91.9 N answer
2] Work done by applied force= F*s*cos theta
= 91.9*4.1*cos 28 degree
= 333 J answer
3] Work done by friction = - 333 J because friction is equal and opposite to horizontal force
4] work done by the normal force = 0 J as there is no displacement in vertical.
5] work done by gravity = 0J
6] Total work done = 0 J
A factory worker pushes a 30.1 kg crate a distance of 4.1 m along a level...