A 3 kg object is initially at rest on a frictionless horizontal surface. A horizontal force of 12 N is applied for 4 s. What is the velocity of the object at the end of the 4 s interval?
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Correct answer
D. 16 m/s
Principle or equation
Newton's second law: F = ma. For constant force, acceleration is constant, and velocity change is a * t.
Why this answer is correct
Acceleration a = F/m = 12 N / 3 kg = 4 m/s^2. Starting from rest, v = a t = 4 * 4 = 16 m/s.
Example
If a 2 kg object is pushed with 10 N for 3 s, a = 5 m/s^2, v = 15 m/s.
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