Work done = 1 N x 1 m = 2 kg 1m2 1.414s-2 = 1 J ??? wrong ??

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1. Jan 7, 2016

Chalmers

• Poster has been reminded to post schoolwork-type threads in the Homework Help forums
~In the vacuum of outer space~

If a rocket motor exerts a force of 1 Newton on a mass of 2 kilograms for 1.414213562 seconds over a distance of 1 meter, how many Joules of work have been done?

I thought
Work = 1 N x 1 m = 1 kg 1m2 1s-2 = 1 J

so
Work = 1 N x 1 m = 2 kg 1m2 1.414s-2 = 1 J ???

I think there is something wrong with my set-up, but what is the proper way to write it?
Is my textbook wrong?
Please see the attached .PNG file for further clarification.

2. Jan 7, 2016

A.T.

"1N per second" is nonsense.

3. Jan 7, 2016

Chalmers

Ok, I think I understand.
Let me re-word my question:

After pushing 2 kilograms a distance of 1 meter, how many Joules of work has a 1 Newton motor (2 kg · 0.5 m/s2) done?

-or-

After running for 1.414213562 seconds, how many Joules of work has a 1 Newton motor (2 kg · 0.5 m/s2) done?

Are these equivalent questions?

And of course, how many Joules?

4. Jan 7, 2016

nasu

And your data is inconsistent. If you accelerate the 2 kg with a force of 1N, you can do it along a distance of 1m OR during 1.4 s.

5. Jan 7, 2016

Staff: Mentor

Yes. Work is force times distance. In both cases, you have 1 N and 1m = 1J.

6. Jan 7, 2016

Staff: Mentor

Am I missing something? Doesn't applying a force of 1N cause a 2kg mass to travel 1m in 1.4s?

7. Jan 7, 2016

HallsofIvy

Staff Emeritus
What you are "missing" is that force causes acceleration not speed. A force of 1 N will cause a 2 kg mass to accelerate at 1/2 m/s^2. How far it will move in that time depends upon the initial speed. Perhaps you are assuming the initial speed was 0 but that was not given.

8. Jan 7, 2016

nasu

Maybe I am missing something.
The acceleration will be $\frac{1N}{2kg} = \frac{1}{2} \frac{m}{s^2}$ , right?
The distance traveled:
$d= \frac{1}{2} a t^2$
$t^2$ is 2 (it looks like he took t as square root of 2).
So $d=\frac{1}{2} (\frac{1}{2} \frac{m}{s^2}) \cdot 2s^2 =1/2 m$
What I am missing?

I think that it needs 2s to travel 1 m, at that acceleration. But I may have a brain failure tonight.

Last edited: Jan 7, 2016
9. Jan 7, 2016

HallsofIvy

Staff Emeritus
Again, you are assuming that this has initial velocity 0- from rest. That was not given.

10. Jan 7, 2016

WrongMan

if initial speed was 0.35m/s... it checks out ..

Last edited: Jan 7, 2016
11. Jan 7, 2016

nasu

Aha, this is how you see it. Thank you for clarification.
It would be interesting to know where that value for time come from.

12. Jan 7, 2016

Staff: Mentor

No - I don't know what I was thinking (I don't remember - either I didn't bother checking the math or did it wrong myself). You're right that they don't match.

13. Jan 8, 2016

Chalmers

1 kg with a 1 N motor would arrive at the 2 meter mark at 1.414 seconds.

14. Jan 8, 2016

Chalmers

Is that correct, though?
I thought that a 1N motor would exert more Joules per amount of time regardless of distance,
as long as we are comparing different masses.
It should take more time (& more Joules) for a 1 N motor to move 10 kg a distance of 1 meter, & less time (& less Joules) to move 1 kg a distance of 1 meter, right?

Last edited: Jan 8, 2016
15. Jan 8, 2016

nasu

It may, if you choose the right initial speed. But not if it start from rest.

The "Joules per second" is power.
P=F*v where F is force and v is magnitude of velocity.
So you either have a motor with constant force and then the power depends on the velocity. So it won't be constant over time if you have accelerated motion.

Or you have a constant power motor and then the force won't be constant unless.

And of course, you can have both changing in time.

The work done is force times distance (W =F*d) so is the same in both cases.
It would be a good idea to use the actual equations instead of trying to put them in words or trying to guess different relationships where a simple relation will show you imediately that your are wrong (or right).

Last edited: Jan 8, 2016