Work, Energy, Power: Accelerating Mass from Rest

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The discussion focuses on calculating the instantaneous power delivered to a body of mass m as it accelerates uniformly from rest to speed v in time T. The initial attempt at a solution used the formula P = w/t, but there was confusion regarding the units of v/T. A suggestion was made to apply the formula P = Fv, leading to a more accurate expression for power as P(t) = m(Vf/T)^2t. The correct approach emphasizes the relationship between force, acceleration, and time to derive the power function. Overall, the conversation highlights the importance of unit consistency and alternative methods in solving physics problems.
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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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