A child pushes a wagon with a passenger of total mass 42 kg along a horizontal surface as fast as the

Question

child can run, and then releases the wagon, which continues for another 16 m before stopping. The coefficient
of kinetic friction acting to slow the wagon is 0.18. What was the speed of the boy when he released the wagon?

Answer 1

work (energy) to stop the wagon
... w = 42 * g * .18 * 16

w equals the initial KE of the wagon
... 1/2 * 42 * v^2 = w

the speed of the boy is the same as the initial speed of the wagon