Asked by Tezara
3.16. An 1100-kg car accelerates from 0 to 30 m/sec in 12 sec. If a frictional force of 200 N opposes the motion, what forces must the wheels exert backward on the pavement to cause this acceleration?
Answers
Answered by
Damon
a = 30/12 = 2.5 m/s^2
F-200 = m a = 1100(2.5) =2750 N
F = 2950 N
F-200 = m a = 1100(2.5) =2750 N
F = 2950 N
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