Engineering Board Exam Reviewer · Engineering Mechanics · medium
A 3.5 kg object's velocity increases from 6 m/s to 11 m/s in 1.25 seconds due to a constant applied force. What is the magnitude of that force?
- 30.8
- 21.88
- 15.12
- 14
Answer
D. 14
Solution
Impulse = p = m(v2 - v1) = 3.5(11 - 6) = 17.5 kg· m/s. Force = Impulse / t = 17.5 / 1.25 = 14 N.
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