Work, Energy, Power: Accelerating Mass from Rest

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SUMMARY

The discussion centers on calculating instantaneous power delivered to a body of mass m, accelerated uniformly from rest to speed v in time T. The correct formula for power as a function of time is derived as P(t) = m(Vf/T)^2t, where Vf is the final velocity. The initial attempt incorrectly used P = [0.5m(v/T)^2] / t, leading to confusion regarding unit consistency. The key takeaway is the application of the formula P = Fv, which simplifies the calculation of power in this context.

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Homework Statement



A body of mass m is accelerated uniformly from rest to speed v in time T. The instantaneous power delivered to the body as a function of time is given by?

Homework Equations



P=w/t

e=1/2mv^2


The Attempt at a Solution



P=w/t
P=[0.5m(v/T)^2 ] /t
p=(mv^2) / (tT^2)


Comments:
The answer given is (mv^2)t / (T^2).
Can anyone tell me where i went wrong?
 
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P=[0.5m(v/T)^2 ] /t

v/T is not in units of m/s. You should always check your units.

However, I would use a different formula, I would use P = F v. Or, more explicitly for this problem P(t) = F v(t)
 
Just a bit of algebraic jugglery. Let Vf be the final velo, and v the instantaneous velo.

P(t) = Fv = mav = ma^2t, since v=at. But Vf = aT. So, a = Vf/T, which gives,
P(t) = m(Vf/T)^2t.
 

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