# MAX POWER question from Dynamics

1. Aug 19, 2007

### CaityAnn

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

A loaded truck weighs 16*10^3 lb and accelerates uniformly on a level road from 15ft/s to 30 ft/s during 4 s. If the frictional resistance is 325 lb determine the max power delivered to the wheels.

2. Relevant equations
I already know the answer, its 119 HP but I dont know how to get to it.

P= F*V = CHANGE WORK/CHANGE IN TIME

since there is no angle here, and the only force is the friction, I found the acceleration using the speed as funtion of time eq, I got A=3.75, they say accelerates uniformly so I assume that means constant acceleration.
Using the speed as a function of position equation I found the change in X to be 90 ft.

i know that KEintial+work due to friction=KE final
So Work = FF*distance, so 325*90, divided by 4 seconds to get power. Then divided by 550 to get to HP.

HELP, Im messing up somewhere, probably with the weight of the truck.:b:yuck:

2. Aug 19, 2007

### mgb_phys

There are two 'forces' the one accelerating the truck and overcoming friction.
F(total) = m a + friction.

Since the friction and accelaration is constant, this force is constant over the distance so you can get the total energy and so power.
How you work it out in those bizarre units is beyond me!

Last edited: Aug 19, 2007
3. Aug 19, 2007

### CaityAnn

thanks for the reply. But your going to have to spell it out for me as I have tried everything known to man.
what does the total energy have to do with it?
I know the mass, times the acceleration found through kinematics, and the fictional force is 375, add it all together to get the total force, then times distance, 90' to get the change in work.
Is that the change in work?
then divide by 4 seconds to get power=change work/change in time, then divide by 550 to convert to HP from ftlb/s.

4. Aug 19, 2007

### mgb_phys

Sounds right,
total force is (mass * acceleration) + friction.
then, Energy is force * distance.
finaly, Power is energy / time.

5. Aug 19, 2007

### CaityAnn

Yes, well, its not working.

6. Aug 19, 2007

### mgb_phys

I get 89.5 HP ( after converting everything to metric) as the average power.

7. Aug 19, 2007

### CaityAnn

me too. Im wondering if the Pmax eq is different, I remember from physics it being slightly different but I cant find a difference.. I have the answer already, it should be 119 HP.

8. Aug 19, 2007

### mgb_phys

Sorry my mistake - that's average power.
But constant acceleration doesn't take constant power, since ke = 1/2 m v^2 as you go faster it takes more power to keep the same rate of acceleration. So the max power is at the final speed.
I get roughly 120 hp ( working in metric and converting )
Have you done calculus yet? if not you can work it out by considering what happens in 0.1 foot at the max speed.

Last edited: Aug 20, 2007
9. Aug 19, 2007

### CaityAnn

yes, i redid everything as maximum speed and got 118.8.
i found the power dissapated= 325*30m/s divded by 550 to convert and got 17.7 HP. Then I found the max energy using the max velocity as 1/2mv^2, converted it to power- 101.6 HP, added both since the power dissipated will need to be included as the intial output power and got 118.8

10. Aug 19, 2007

### CaityAnn

thank you also.