# Homework Help: Calculating power input

1. Apr 6, 2013

### mouthwash

A 1200kg car goes a distance of 12 meters on a road, going from 4m/s to 18 m/s in 6 seconds. If 75,000J of energy is used as heat energy, calculate power in/out and efficiency.

I believe I found the correct power output, by doing (.5)(1200)(14^2)/6 = 19600W, but I dont know how to find the power input. Would I just add in the energy lost to friction to the power output? I tried this (19600W /32100W * 100% = 61%) but according to the answer book this is the wrong answer.

2. Apr 6, 2013

### cepheid

Staff Emeritus
Welcome to PF, mouthwash!

(removed boldface because it is annoying :tongue:)

I don't think your power output is correct, unfortunately. The power is given by P = ΔE/Δt, the change in energy over the change in time. So you are correct about that. In this case, the energy is kinetic, associated with the car's motion. So ΔE would be ΔKE, which would be:

ΔKE = KEfinal - KEinitial

which would in turn be:

(1/2)m(vfinal)2 - (1/2)m(vinitial)2

Factoring things out, we'd end up with:

ΔKE = (1/2)m[ (vfinal)2 - (vinitial)2 ]

Look at the quantity in square brackets. It's not the same thing as what you wrote. You wrote (vfinal - vinitial)2, which is different, and wrong. You have to square the speeds first, and then subtract them. Instead, you subtracted the speeds, and then squared the result.

3. Apr 6, 2013

### mouthwash

okay if I do it like that, I get answer for power output of 12800, but i still recieve the wrong answer when i do 12800/(12800*6 + 75000) * 100% = 51%. The answer should be 71% for efficiency.

4. Apr 6, 2013

### cepheid

Staff Emeritus
Also, the problem tells you that 75,000 J of energy from the engine ends up in the form of heat. So, a certain amount of energy is provided by the engine (over six seconds) but only some of that goes into useful work (which is the change in kinetic energy that you computed). The rest is wasted as heat. So the energy input would just be (useful work + waste heat). You can figure out the efficiency just from (energy out)/(energy in).

5. Apr 6, 2013

### cepheid

Staff Emeritus
There are a couple of problems.

1. 12,800 W is the wrong answer for the power output. Try again.

2. You're mixing power and energy in your efficiency calculation (which is why you should ALWAYS include units in your computations, to avoid mistakes like this). EITHER use

efficiency = (power out)/(power in)

OR use

efficiency = (energy out)/(energy in)

It doesn't matter, because the second equation is just the first equation multiplied by (6 seconds)/(6 seconds) = 1.

6. Apr 6, 2013

### mouthwash

I see, thanks a lot. I got the right answer finally.