Power of constant force for a given displacement

AI Thread Summary
The discussion focuses on calculating the power developed by a constant force acting on an object moving from y = 0 to y = 1.5 m with a final velocity of 8 m/s. Two approaches are explored: using P = Fv for instantaneous power and P = W/t for average power. The initial confusion arises from treating velocity as constant, which is incorrect since it changes during motion. The correct method for instantaneous power requires using P = dW/dt, emphasizing the need for accurate velocity representation throughout the motion. The conversation clarifies the distinction between instantaneous and average power calculations.
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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