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The power output of a motor acting on a 2 kg body is given by P = v*(5 - v), where v is the speed in m/s. The body starts from rest. Find the work done on the body in the first 2 seconds.

1. 25(e - 1)
2. 5((e - 1)/e)
3. 25((e - 1)/e)²
4. 25((e² - 1)/e²)²

Correct answer: 25((e - 1)/e)²

Solution

From F = 5 - v and m = 2: 2*(dv/dt) = 5 - v => dv/(5-v) = dt/2. Solving: ln(5/(5-v)) = t/2 => v = 5(1 - e^(-t/2)). Then P = v(5-v) = 5(1-e^(-t/2)) * 5*e^(-t/2) = 25*e^(-t/2)*(1-e^(-t/2)). W = integral from 0 to 2 of 25*e^(-t/2)*(1-e^(-t/2)) dt = 25*integral(e^(-t/2) - e^(-t))dt from 0 to 2 = 25*[-2e^(-t/2) - (-e^(-t))] from 0 to 2 = 25*[(-2/e + 1/e²) - (-2 + 1)] = 25*[1 - 2/e + 1/e²] = 25*(1 - 1/e)² = 25*((e-1)/e)².

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