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?

Answer 1

Kinetic energy gained during the frictionless interval is 400 N * 17 m = 6800 J. When there is friction, the net force pulling the weight is 400 N - M*g*mu = 400 - 399.8 N = 0.2 N. Additional kinetic energy gain is limited to 0.2N*17m = 3.4 J

The rest of the work done ends up as heat.

Final kientic energy is 6803 J. Use that for the final velocity

Answer 2

how would I show that in m/s?

Answer 3

you use 1/2*m*v^2

Answer 4

the answer is = 1