A 3.0 kg object is initially at rest on a frictionless horizontal surface. A horizontal force of 15 N is applied for 6.0 s. What is the velocity of the object at the end of the 6.0 s interval?
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Correct answer
C. 30 m/s
Principle or equation
Newton's second law: F = ma, and kinematics for constant acceleration: v = v0 + at.
Why this answer is correct
First find acceleration: a = F/m = 15 N / 3.0 kg = 5 m/s². Then use v = v0 + at. Since v0 = 0, v = 5 m/s² × 6.0 s = 30 m/s.
Example
If a 2 kg object is pushed with a 10 N force for 3 s from rest, a = 5 m/s² and v = 15 m/s.
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