A 120-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 400 N. For the first 17 m the floor is frictionless, and for the next 17 m the coefficient of friction is 0.34. What is the final speed of the crate? Express your answer using two significant figures and m/s.

Question

A 120-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 400 N. For the first 17 m the floor is frictionless, and for the next 17 m the coefficient of friction is 0.34.   What is the final speed of the crate? Express your answer using two significant figures and m/s.

drwls · Answer

Final kinetic energy = (work done) - (work lost due to friction)
(1/2)M*Vf^2 = 400N*34 m - M*g*Uk*(17 m) = 13,600 J - 6797 J
Solve for Vf.

There is almost no acceleration during the final 17 m. The pulling force nearly equals the friction force. This is a coincidence; they set it up that way