A 3.0 kg object is initially at rest on a frictionless horizontal surface. A horizontal force of 12 N is applied for 4.0 s. What is the velocity of the object at the end of the 4.0 s interval?
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Correct answer
C. 16 m/s
Principle or equation
Newton's second law: F = ma gives acceleration. Then v = v0 + at with v0 = 0.
Why this answer is correct
a = F/m = 12 N / 3.0 kg = 4.0 m/s². Then v = 0 + (4.0 m/s²)(4.0 s) = 16 m/s.
Example
A 2.0 kg object has a force of 10 N applied for 5 s. a = 5 m/s², v = 25 m/s.
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