A student is pushing a crate with a constant force of 40 N directed at 20 degrees below the horizontal. The mass of the crate is 20 kg. The crate is moved by 15 m. Determine the coefficient of kinetic friction, work done by the student and work done by the frictional force.
1 Answer | Add Yours
The student exerts a force of 40 N on the crate, the force is directed at an angle 20 degrees below the horizontal. This results in the crate moving at a constant velocity for a distance 15 m.
We have to determine the coefficient of kinetic friction. The horizontal component of the force applied by the student on the crate is 40*cos 20. The force of friction is n*20*9.8 where n is the coefficient of kinetic friction.
n*20*9.8 = 40*cos 20
=> n = 40*cos 20/(20*9.8)
=> n = 0.1917
The coefficient of kinetic friction is 0.1917.
The work done by the student in moving the crate is 40*cos 20*15 = 563.81 J. The work done by the force of friction is also 563.81 J.
We’ve answered 318,957 questions. We can answer yours, too.Ask a question