Calculating power as a function of time

[hk4491]

Homework Statement

A force of 5 N acts upon a body of 8 kg in the +x direction. Formulate an expression for the power generated as a function of time. The body is in the beginning at t=0 and x=0 (Sorry I'm translating this question from German, so excuse the mistakes).

Homework Equations

Work = Force * distance

Power = ΔW/Δt

Force = mass * acceleration

The Attempt at a Solution

Since the force of 5N is acting on the body, by Newton's third law, it is also exerting the same force, so that:

F = m * a

5 = 8*a

a=5/8 m/s^2

By integrating the acceleration twice I get the distance as a function of time:

x(t) = 5/8 t^2

Work = Force * distance = 5 * (5/8)t^2 = (25/8)t^2

P(t) = (25/8)t

Was my method correct?

 

[haruspex]

By integrating the acceleration twice I get the distance as a function of time:

x(t) = 5/8 t^2

You forgot something in doing that integration.

Work = Force * distance = 5 * (5/8)t^2 = (25/8)t^2

P(t) = (25/8)t

Was my method correct?

You have divided the total work done over the distance by the total time taken. That will give you average power, but I think it is the instantaneous power at time t that is wanted.

 

[hk4491]

Thanks I realized my mistake in the integration part. That leads to x(t)= 5/16 t^2.

The body starts moving at t0=0, so at any given point of time Δt = t - t0 = t

Hence:

P= (F*d)/t

P(t) = 25/16 t

 

[Curious3141]

Thanks I realized my mistake in the integration part. That leads to x(t)= 5/16 t^2.

The body starts moving at t0=0, so at any given point of time Δt = t - t0 = t

Hence:

P= (F*d)/t

P(t) = 25/16 t

No. You missed out a factor of half in your integration, so the correct expression for total work done at time t should be $W(t) = \frac{25}{16}t^2$. This expression is not linear in t, so can you simply divide by t to find the power at time t? Shouldn't you be differentiating?

Basically, your original expression was correct - simply because your errors "cancelled out". You missed a half when integrating, then divided instead of differentiating, so you didn't include a requisite factor of 2.

 

[hk4491]

Great, thanks for the explanation!

 

[Chestermiller]

Another way of arriving at the answer is to remember that the instantaneous power is equal to the force times the velocity. The velocity at time t is 5t/8.