A 2.0 kg object is moving across a table at a constant speed of 2.0 m/s against a force of friction of 6.0 N.The applied force is suddenly increases by 1.0 N. How fast will the object be moving after 3.0s?

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goformit100 · Answer

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anonymous · Answer

Vf = Vi + at
Initially the object is moving at a constant velocity, so we know that all forces are balanced. As friction does not change with velocity (ignoring air resistance and all that silly stuff), we can calculate the acceleration of the object using f = ma, where f is 1.0N, m is the mass of the object, and a is the desired acceleration.

rearranging we get a = f/m, which is a = 1N / 2kg = 0.5 ms^-2

Now we have a, we can plug it all into Vf = Vi + at
where Vi is the initial velocity, 2.0 m/s, a is the acceleration we just calculated 0.5 ms^-2 and t is the time given, 3 seconds. 

this give Vf (final velocity after 3 seconds) = 2 m/s + 0.5*3 = 3.5 m/s

anonymous · Answer

Thanks so much peterorsome!