The power delivered by a motor to a body of mass 2 kg is given by P = v*(5 - v) watts, where v is the speed in m/s. If the body starts from rest, find the work done on the body during the first 2 seconds.

25*(e - 1) J
5*(e - 1)/e J
25 J
25*((e - 1)/e)² J

Correct answer: 25*((e - 1)/e)² J

Solution

From F = 5-v = 2*(dv/dt), solving gives v(2) = 5*(1 - e⁻¹) = 5*(e-1)/e. Work done = Delta KE = (1/2)*2*v² = v² = 25*((e-1)/e)².

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