A 2 kg object moves under the influence of a force such that its displacement is given by x = t³ / 3, where x is in metres and t is in seconds. Find the work done on the object during the first 2 seconds.

8 J
16 J
24 J
32 J

Correct answer: 16 J

Solution

Given x = t³/3. Velocity v = dx/dt = t². At t=0: v=0. At t=2: v=4 m/s. By the work-energy theorem: Work = delta(KE) = (1/2)*m*v² - 0 = (1/2)*2*(4)² = (1/2)*2*16 = 16 J.

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