Power of car going up an inclined road

1. Oct 13, 2013

coconut62

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

Please refer to the image attached.

2. Relevant equations

P=Fv

3. The attempt at a solution

I have solved the problem already.
But when I refer to the marking scheme, I found that they don't accept the use of P=Fv.
I understand it's because there's an acceleration, but what if I take the average velocity?

Is that valid?

Attached Files:

• 1376260_10151784811557830_2039627177_n.jpg
File size:
45.6 KB
Views:
65
2. Oct 13, 2013

Tanya Sharma

Can you write the original problem as given to you? What are the given parameters ?

3. Oct 13, 2013

coconut62

The value provided are all correct. They came from previous calculations which ask for the time taken/Ke and Pe gained etc.

4. Oct 13, 2013

Simon Bridge

What they probably wanted was $P=\Delta E/\Delta t$ ... since they asked about the time it took.
We need to see stuff like the nature of the acceleration in order to advise you properly.

5. Oct 14, 2013

coconut62

Okay, here is the full question:

Attached Files:

File size:
45.4 KB
Views:
61
File size:
29.8 KB
Views:
58
• 1380693_10151790102182830_1123361028_n.jpg
File size:
34.8 KB
Views:
63
6. Oct 14, 2013

Tanya Sharma

Yes . Pavg=Fvavg.

Last edited: Oct 14, 2013
7. Oct 14, 2013

Simon Bridge

The instructions say
(ii) Use your answers in (i) to determine the useful output power of the car.

You didn't do this.
You were penalized, in effect, for not following instructions.
The question was, indeed, testing you to see if knew that $P=\Delta E /\Delta t$ ... but you did not demonstrate that you knew this because you used a different, more convoluted, method.
So the marker was unable to award you with some marks.

You can answer your own question by comparing the answer using the average velocity with the one using the work-energy relation.

Last edited: Oct 14, 2013
8. Oct 14, 2013

MrAnchovy

Actually coconut62 did use $P=\Delta E /\Delta t$, and I assume got 3 marks for this section (or possibly 2 if a mark was lost for not stating the equation).

Otherwise Simon your point is valid - the marking scheme does not give credit for P= Fv because you didn't work out F or v in part 1, you worked out ΔE and Δt.

Last edited: Oct 14, 2013
9. Oct 14, 2013

Simon Bridge

<closer look at attachments> ... actually, in the material provided, "P=Fv" appears to be mentioned only by the marker ???

10. Oct 16, 2013

coconut62

^ Yes, because some people would use P=Fv and the marker don't want to give mark for that. At first I thought it's invalid because P=Fv would give a wrong answer, because I didn't notice the ~instruction~. lol

11. Oct 16, 2013

imiuru

Now that you know it is the instruction that's the problem, would you be able to solve it using P=Fv ?

12. Oct 17, 2013

coconut62

I got a miserable answer. Not even close.

Attached Files:

• 1376627_10151796246122830_1690228250_n.jpg
File size:
40 KB
Views:
49
13. Oct 17, 2013

Simon Bridge

Box mass m going distance L up a slope angle θ to horizontal, from rest to speed v, in time T, gains energy

$E=KE+PE=\frac{1}{2}mv^2+mgL\sin\theta$

...in that time so the power comes to

$P=\frac{1}{2}mv^2/T+(mg/T)\sin\theta$ ...(1)

The other way: $P=F\bar{v}$ using $F=ma$ and $\bar{v}=v/2$

$P= m(v/T)(v/2) = \frac{1}{2}mv^2/T$ ...(2)

... compare this result with the first one... what's missing?
Therefore - when can you use the formula P=Fv?

14. Oct 17, 2013

coconut62

I think in (1), your L is missing.

-----
Okay, so the PE part is missing.
Because PE doesn't involve velocity, so P=Fv can't be used here. (?)

But since the box is going up the slope with an acceleration, which means that its rate of change of PE is also increasing, which means there is a velocity(somewhere), then why can't I take the components?

15. Oct 17, 2013

Simon Bridge

That was just a um...
test...
that's right, to see if you were paying attention....
and you were... um... well done :)

In this particular case P=Fv does not take int account the change in potential energy - only change in kinetic energy.

Lets see ... $\vec{v}=v_x\hat{\imath}+v_y\hat{\jmath}$ then in terms of components:
$P=\frac{1}{2}m(v_x^2+v_y^2)/T+mgv_y$ ... where does that get you?

Lets try a simpler setup:

Lets say the box is just lifted straight upwards through a height h at a constant speed v, so the task is completed in time T=h/v.

What is the power expenditure by the different formulas:

P=Fv=mgv

P=E/T=mgh/T=mgv

... all the calculations agree.

Now we add some constant acceleration - so the box is lifted through height h, but the initial speed is u and the final speed is v ... as well as the mgh gained, the box also gains some additional kinetic energy.

Using the formula P=Fvave
F=ma=m(v-u)/T
vave=h/T=(v-u)/2

$P=m(v-u)^2/2T = \frac{1}{2}m(\Delta v)^2/T$

... if u=0 then Δv=v and that is the kinetic energy contribution you saw before.

Using conservation of energy:

$P=E/T = \frac{1}{2}m(v^2-u^2)/T+mgh/T$

Note: off the P=Fv result -
$$\frac{1}{2}m(v-u)^2=\frac{1}{2}m(v^2-u^2) + \frac{1}{2}m(u^2+u^2-2uv)$$
... which is the kinetic energy term and another one.

The two methods are the same if the second term is potential energy.
This happens if:

$gh = u^2-uv$

... is it? :)
Clearly not in every case ... i.e. when u=0.