Power of constant force for a given displacement

[sunquick]

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.

 

[jackarms]

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.

 

[dauto]

In the formula P = Fv, if you want to calculate the average power than you ought to use the average velocity. Did you?

 

[sunquick]

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?

 

[dauto]

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