A force of 30 N accelerates a 2.0 kg object from rest for a distance of...

Question

Question

A force of 30 N accelerates a 2.0 kg object from rest for a distance of 3.0 m along a level, frictionless surface; the force then changes to 15 N and acts for an additional 2.0 m. (a) What is the final kinetic energy of the object? (b) How fast is it moving?

Answer 1

Answer #1

Answer:

(a) 120.0 J

(b) 10.95445 m/s

We know that,

work done = (force applied) * (displacement of point of application of force )

Work done during first 3.0 m = w1 = (30 N )*(3.0 m) = 90.0 J

Work done during next 2.0 m = w2 = (15 N) * (2.0 m) = 30.0 J

Change in kinetic energy = total work done = w1 +w2 = 90.0 J + 30.0 J = 120.0 J

Also, initial kinetic energy = 0 ( mass was at rest)

So final kinetic energy = 120.0 J

And, kinetic energy = (1/2)*m*v2 = 120.0 J

Or, (1/2)*2 kg * v2 = 120.0 J

Or,

Answer 2

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