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Question

A 4.5 kg object is accelerated from rest to a
speed of 39.2 m/s in 76 s.
What average force was exerted on the ob-
ject during this period of acceleration?
Answer in units of N
13 years ago

Answers

Henry
a = (Vf - Vo) / t,
a = (39.2 - 0) / 76 = 0.516m/s^2.

F = ma = 4.5 * 0.516 = 2.32N.
13 years ago

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