Engineering Board Exam Reviewer · Engineering Mechanics · medium
A 2.5 kg object's velocity increases from 4 m/s to 7 m/s in 0.75 seconds due to a constant applied force. What is the magnitude of that force?
- 23.33
- 10
- 5.63
- 10.8
Answer
B. 10
Solution
Impulse = p = m(v2 - v1) = 2.5(7 - 4) = 7.5 kg· m/s. Force = Impulse / t = 7.5 / 0.75 = 10 N.
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