Question

A 3.1 kg object is accelerated from rest to a
speed of 61.6 m/s in 66 s.
What average force was exerted on the ob-
ject during this period of acceleration?
Answer in units of N

Answer 1

v2 = v1 + at
61.6 m/s = 0 + a(66s)
a =61.6/66
a= 14/ 15 m/s^2

F=ma
F=(3.1kg)(14/15)
F=2.893 N

Answer 2

The answer by Riskydiablo is certainly correct but remember that the answer should in practice not be quoted to more decimal places than the least accurate data value. In this case the the answer should be given as 3 N.

Another way of getting the answer is to remember that when a question talks about average forces you can think in terms of the following equation\[F _{av}=m \Delta a=m (\Delta v/\Delta t) \] where\[ \Delta v/\Delta t) = (v_{f}-v_{i})/(t_{f}-t_{i})\] In this case \[(v_{f}-v_{i})/(t_{f}-t_{i})=(61.6-0)/66-0)\]