Initial and final power supplied by a person pushing a stone

[Howard Fox]

Homework Statement

Homework Equations

The Attempt at a Solution

I solve question a) by rearranging the F=m*a equation, which makes the acceleration equal to (24N/18kg)=1.30m/s^2. And I find the final speed using equation W=1/2*m*v^2 and rearranging it to solve for v, which gives me a value of v=3.27m/s.
But the part of the question I am having trouble with is when it asks me to find how much power is supplied at the beginning and end of the push. How to find that? I know that average power equals change in work over change in time, but using this formula doesn't seem to lead anywhere. I guess there must be another equation I can't think of?

Anybody has a better attempt?

 

[gneill]

Did you try googling "kinematic equation for power"?

By the way, your "Relevant equations" entry is really your attempt at solution. I'll move it.

 

[Howard Fox]

kinematic equation for power

Okay, thank you.
And what are the kinematic equations for power? You mean one these 3? :

I don't know of anything else and google can't seem to help me on this.
Thank you for your help

 

[gneill]

And what are the kinematic equations for power? You mean one these 3?

Yes!
Take a good look at the one in the box

 

[Howard Fox]

Yes!
Take a good look at the one in the box

You mean the last one? Power equals force times velocity? I tried working with that one, but how would I find the initial and final power? Should I put v equal to 0 for the initial and Vfinal for the final? But that would imply that the initial power is equal to zero, which doesn't make sense to me intuitively. I could think reason though!

 

[gneill]

You mean the last one? Power equals force times velocity? I tried working with that one, but how would I find the initial and final power? Should I put v equal to 0 for the initial and Vfinal for the final?

Yes.

But that would imply that the initial power is equal to zero, which doesn't make sense to me intuitively. I could think reason though!

If the stone hasn't moved yet, no energy has been transferred. Hence no power has been spent.

edit: removed spurious QUOTE tag.

 

[Howard Fox]

Yes.

If the stone hasn't moved yet, no energy has been transferred. Hence no power has been spent.

[/QUOTE]
Okay, thank you for the explanation
So the initial power is equal to zero and the final power is the force times the final velocity, which gives 78.48W. But this still doesn't seem to give me a way to answer the last question. I know that the time of the push is found by using the standard kinematic equation: V=Vo+a*t and solving for t, which gives me a value of 2.5s. But what about the second part of the last question? The numbers don't seem to add up

 

[gneill]

I know that the time of the push is found by using the standard kinematic equation: V=Vo+a*t and solving for t, which gives me a value of 2.5s. But what about the second part of the last question? The numbers don't seem to add up

What have you tried?

 

[Howard Fox]

What have you tried?

I found the average power by dividing the product of the final power and initial power by 2, which gives me average power=39.24w.
Then I rearranged the equation avg.Power=(F*Sx)/t, solving for t, which gives me 96J/39.24W=2.44s. Slightly different from the time I calculated in the other equation
Maybe I am doing this all wrong

 

[haruspex]

dividing the product of the final power and initial power by 2

You mean the sum, not the product.
But the question is a bit strange here. It hadn't asked you to find the average power, implying the questioner thinks that finding it is trivial. Average power over some interval is the integral of power over the interval divided by the duration of the interval. Yet the question asks you to find that duration from the average power!
So my guess is you have done what the questioner wanted, but the question should not have called it "average power".

 

[Howard Fox]

You mean the sum, not the product.
But the question is a bit strange here. It hadn't asked you to find the average power, implying the questioner thinks that finding it is trivial. Average power over some interval is the integral of power over the interval divided by the duration of the interval. Yet the question asks you to find that duration from the average power!
So my guess is you have done what the questioner wanted, but the question should not have called it "average power".

okay, complicated! So what do you think I should answer in the second part of the last question? I mean I should find some value of which equals 2,52s and I can't find a way to find one!

 

[gneill]

dividing the product of the final power and initial power by 2, which gives me average power=39.24w.

You should show us how you calculated the final power. Your 39.24 W value for its half seems a bit high. This could be where your slight difference in time is coming from .

I mean I should find some value of which equals 2,52s and I can't find a way to find one!

Where does the 2.52 s figure come from?

 

[Howard Fox]

You should show us how you calculated the final power. Your 39.24 W value for its half seems a bit high. This could be where your slight difference in time is coming from .

I just divided that number by two to get 39.24W
Where does the 2.52 s figure come from?

 

[gneill]

Ah, the perils of rounding intermediate values! If I don't round at intermediate steps and keep more digits I see:

Final velocity: $v = 3.266\;m/s$
Final Power: $P_f = 78.384\;W$
Elapsed time: $t = 2.449\;s$

Try your calculations again without rounding intermediate values.

 

[Howard Fox]

Ah, the perils of rounding intermediate values! If I don't round at intermediate steps and keep more digits I see:

Final velocity: $v = 3.266\;m/s$
Final Power: $P_f = 78.384\;W$
Elapsed time: $t = 2.449\;s$

Try your calculations again without rounding intermediate values.

I don't understand. This is still not equal to the time I found using this equation:

Even with more precise rounding, I still get t=2.512s
The two values for time should be the same, no?

 

[haruspex]

24N/18kg)=1.30m/s^2

Try that again.

 

[gneill]

I don't understand. This is still not equal to the time I found using this equation:

Even with more precise rounding, I still get t=2.512s
The two values for time should be the same, no?

You are using too-rounded results again.

$V_f = 3.266\;m/s$
$a = 1.333\;m/s^2$
$V_f/a = 2.449 s$

 

[gneill]

But the Vf is wrong because the acceleration it was based on was wrong.

True! I had (mistakenly) presumed that 1.333 had been rounded to 1.3 . But now that I look at the initial post I see that it was written as 1.30. So yes, the value was incorrect from the start.

 

[Howard Fox]

True! I had (mistakenly) presumed that 1.333 had been rounded to 1.3 . But now that I look at the initial post I see that it was written as 1.30. So yes, the value was incorrect from the start.

Okay, I think I got it now. The thing is when I put 24/18 in my calculator it gives me 1.3, not 1.333. That's the origin of the confusion I guess

 

[haruspex]

okay, complicated! So what do you think I should answer in the second part of the last question? I mean I should find some value of which equals 2,52s and I can't find a way to find one!

Let me clarify...
I am saying that, in general, taking the average of initial and final power is not a legitimate way to find average power. It happens to work in this case because the force is constant, so velocity increases uniformly with time, and so power also increases uniformly with time.
I was concerned that (when you finally get the two numbers to match) the question might leave the solver with the impression that this method works in general.

 

[haruspex]

True! I had (mistakenly) presumed that 1.333 had been rounded to 1.3 . But now that I look at the initial post I see that it was written as 1.30. So yes, the value was incorrect from the start.

I deleted my post because I then realized the later numbers were correct. Posting the acceleration as 1.30 seems to have been a transcription error. Maybe Howard continued the calculation using the number actually in the calculator, not the number being displayed.

 

[Howard Fox]

I deleted my post because I then realized the later numbers were correct. Posting the acceleration as 1.30 seems to have been a transcription error. Maybe Howard continued the calculation using the number actually in the calculator, not the number being displayed.

Wait, so 1,333 is not valid? Because if I use 1.3 the different values of t don't match. But if I use 1.333 they match.

 

[gneill]

Wait, so 1,333 is not valid? Because if I use 1.3 the different values of t don't match. But if I use 1.333 they match.

It is valid.

 

[Howard Fox]

It is valid.

Why doesn't my calculator give me the same value?

 

[gneill]

Why doesn't my calculator give me the same value?

Now that I couldn't say. You'll have to look into the instruction manual for your calculator to see if there are settings for rounding results to a certain number of figures.

 

[haruspex]

Wait, so 1,333 is not valid? Because if I use 1.3 the different values of t don't match. But if I use 1.333 they match.

No, no! I meant that it looks like your calculator had correctly calculated 1.333.., however many digits, and had used that in subsequent steps in your calculation, but, for some reason, it had only displayed as 1.3.
That does seem weird. Perhaps you should experiment with it.

Edit: a possibility.. does it put a little dot over the 3?

 

[Howard Fox]

No, no! I meant that it looks like your calculator had correctly calculated 1.333.., however many digits, and had used that in subsequent steps in your calculation, but, for some reason, it had only displayed as 1.3.
That does seem weird. Perhaps you should experiment with it.

Edit: a possibility.. does it put a little dot over the 3?

No, my calculator gave me the number 1.3 because it wasn't set to linear calculations. It gave me fractions!