Asked by Jatin
A child pushes a wagon with a passenger of total mass 42 kg along a horizontal surface as fast as the
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?
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?
Answers
Answered by
Scott
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
... 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
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