Question

An initially stationary 4.1 kg object accelerates horizontally and uniformly to a speed of 9.4 m/s in 3.6 s. (a) In that 3.6 s interval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval?

Answer 1

a)

workdone = change in KE

W = 0.5* m ( v^2 - u^2)

W= 0.5* 4.1 * ( 9.4^2- 0)

W = 181.14 J

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b)

Acceleration of the object

a = ( v-u) / t = 9.4 / 3.6 = 2.611 m/s^2

P = F v

P = ma v

P = 4.1 * 2.611* 9.6 = 102.77 W

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c)

Power at t = 1.8

P' = P/2 = 51.38 W

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