# Power of constant force for a given displacement

1. May 11, 2014

### sunquick

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$v^2 = 2 a y$

$v = a t$

$P = F v$

$P = \frac{W}{t}$

3. 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.

2. May 11, 2014

### 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.

3. May 11, 2014

### dauto

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

4. May 12, 2014

### 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?

Last edited: May 12, 2014
5. May 12, 2014

### dauto

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