Power of constant force for a given displacement

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Homework Help Overview

The problem involves calculating the power developed by a constant force acting on an object moving from a position of y = 0 to y = 1.5 m, starting from rest. The final velocity at y = 1.5 m is given as 8 m/s. The discussion revolves around the application of various power equations and the implications of using average versus instantaneous values.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss two approaches to calculating power: one using the final velocity and another using work over time. There is a noted paradox in the original poster's calculations, leading to questions about the validity of the methods used.

Discussion Status

Some participants have provided feedback on the approaches taken, suggesting that the first method may not account for the changing velocity throughout the motion. Others have raised the question of whether the second approach can also be used for instantaneous power calculations, indicating a productive exploration of the topic.

Contextual Notes

There is an emphasis on distinguishing between average and instantaneous power, with participants questioning the assumptions made in the calculations. The original poster expresses uncertainty about the validity of their methods and seeks clarification on the appropriate use of formulas.

sunquick
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Homework Statement


An object is subject to a constant force of known magnitude, starting the motion from y = 0 , and with zero initial velocity. Calculate the power developed by the force during the motion of the object from y= 0 to y = 1.5, and at y=1.5m , v = 8 m/s.


Homework Equations


v^2 = 2 a y

v = a t

P = F v

P = \frac{W}{t}

The Attempt at a Solution


P = F v = F \sqrt{2ya}

but on the other hand
t = \frac{v}{a} = \frac{v}{v^2/2y} =\frac{2y}{v}

P =\frac{W}{t} =\frac{Fy}{t} = \frac{F}{2} v = \frac{F}{2} \sqrt{2ya}

So I tried working the problem out in two different ways, and I get a paradox:
F = \frac{F}{2}

I must have done something stupid like dividing by zero but I can't figure out really where I messed up.
 
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Your second approach is correct -- the issue with the first one is that the velocity isn't constant, so if you use the final velocity it doesn't take into account the earlier changes in velocity.
 
In the formula P = Fv, if you want to calculate the average power than you ought to use the average velocity. Did you?
 
Thank you for the replies.

I was trying to calculate the power at the end of the motion, y=1.5 m . At that instant I can calculate the power, since I care only for the final velocity, which I know, then P = F v.
The second approach P = W/t is valid in this case to calculate the instantenous power too right?

The average speed is

v_{avg} = \frac{y}{t}

t =\frac{v_{f}}{a}

y=\frac{v_{f}^2}{2a}

v_{avg} = \frac{v_{f}}{2}

v_{f}^2= 2ay (f for final)

So my first approach resulted in the instanteneous power at the end of the motion, while the second resulted in the average power during all of the motion.

I still have one question: The second approach P = W/t is valid in this case to calculate the instantenous power too?
 
Last edited:
sunquick said:
Thank you for the replies.

I was trying to calculate the power at the end of the motion, y=1.5 m . At that instant I can calculate the power, since I care only for the final velocity, which I know, then P = F v.
The second approach P = W/t is valid in this case to calculate the instantenous power too right?

The average speed is

v_{avg} = \frac{y}{t}

t =\frac{v_{f}}{a}

y=\frac{v_{f}^2}{2a}

v_{avg} = \frac{v_{f}}{2}

v_{f}^2= 2ay (f for final)

So my first approach resulted in the instanteneous power at the end of the motion, while the second resulted in the average power during all of the motion.

I still have one question: The second approach P = W/t is valid in this case to calculate the instantenous power too?

No, if you want instantaneous power you have to use P = dW/dt
 

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