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

Tags:
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