a 3.6 kg object is accelerated from rest to a speed of 34.4 m/s in 42 s. What average force was exerted on the object during this period of acceleration. Answer in units of N.

Answer 1

F = m*a
a = V-Vo)/t = (34.4-0)/42 = 0.819 m/s^2.

F = 3.6*0.819 = 2.95 N.