Question

A 0.100μg speck of dust is accelerated from rest to a speed of 0.900 c by a constant 1.00×10^6N force.

Answer 1

If the nonrelativistic form of Newton's second law F=ma is used, how far does the object travel to reach its final speed?

Answer 2

@theEric Hey, do you know how to do this?

Answer 3

Hi! Well, if your not taking special relativity into account, you can use fundamental kinematic equations!
You know the mass, force, and initial and final velocity.

From the force and mass, you can get the acceleration.

You can skip to a derived equation with acceleration, initial and final velocity, and displacement to answer this.

To go on to applying special relativity, I would need a refresher.

Answer 4

This involves some sort of relativity but I have no idea what, I see you can get acceleration but how would you go on to get the distance travelled in its final speed?

Answer 5

\(v^2=u^2+2a\Delta x\)
I forget how to derive that :P

Answer 6

Ah so, \[\Delta x = \frac{ v^2 }{ 2a } \implies \frac{ 0.900c }{ 2(9.23*10^{12}m/s^2) }\]

Answer 7

Sorry the acceleration is wrong, wrong numbers xD

Answer 8

a = 1*10^13 m/s^2

Answer 9

Oh and that should be squared...(0.900c)^2 bleh

Answer 10

I'll agree after you also square the numerator to make - oh, you got it :)

Answer 11

Yep. That looks good for classical physics as far as I know!

Answer 12

\[0.100 \mu g = 0.100*10^{-6} kg?\] right because I think I did something wrong, not sure what..mhm

Answer 13

Ah, that's the issue! Sorry I didn't catch it either! It's just grams, there. So it's \(10^{-3}\) kilograms.

Answer 14

uh, that's the order, not the number.

Answer 15

micrograms isn't it haha? I need to go look at the metric system I thought it was -6

Answer 16

\(0.1\times10^{-3}=10^{-4}\)

Answer 17

It is! \(10^{-6}\) grams, though!

And I messed up!

Answer 18

\(0.1\times10^{-9}\) grams

kg
g
mg
\(\mu\)g

Answer 19

oh yeah it's something like 10^-10 then lol

Answer 20

One sec, let me use my name to make sure!

Answer 21

Haha, good plan.

Answer 22

http://www.wolframalpha.com/input/?i=0.100+micrograms+in+kg

Answer 23

I thought it was ^-10 kg

Answer 24

Yup there we go haha

Answer 25

Haha, yep! Your name agrees.

Answer 26

Alright, there's a second part mind helping me with it, I'm not really sure what K means here...Using the correct relativistic treatment (K=(y-1)mc^2) , how far does the object travel to reach its final speed?

y=gamma

Answer 27

Or is K suppose to be kinetic energy

Answer 28

\[\gamma = \frac{ 1 }{ \sqrt{1-v^2/c^2} }\] I guess to find gamma we could use this, but not sure about K and the distance.

Answer 29

I think that might be relativistic kinetic energy!

Answer 30

And \(\gamma\) is like the Lorentz transformation or something, right?

Answer 31

Yeah so wait a sec, we have this then \[K = \frac{ mc^2 }{ \sqrt{1-v^2/c^2} }-mc^2 = (\gamma - 1)mc^2\]

Answer 32

So I can get gamma and figure out K but then for distance?

Answer 33

Good question...

Answer 34

The \(\gamma\) will change with the velocity...

Answer 35

Can we use work at all somehow? The force is doing work, right? But I don't know how to deal with special relativity!

Answer 36

I'm sure we can since \[K = W = \int\limits_{0}^{v} \frac{ mv_x dv_x}{ (1-v_x ^2/c^2)^{3/2} }\]

Answer 37

That's how you would derive the expression above.

Answer 38

\[W = Fd ?\] can we just use that? Or do we have to integrate

Answer 39

W = K so the expressions for K/F = d

Answer 40

This is a pretty cool question as far as comparing classical and relativistic aspects will be neat to see the final answer on which one has a greater distance.

Answer 41

Energy is conserved, even in special relativity, right?
So to have a gain in kinetic energy that you calculate with your equation, that much work has to be done. It's done by that given force.

So I think it is that simple! :)

Answer 42

I agree! :)

Answer 43

Since we're looking at two points only, I don't think integration is necessary :)

Answer 44

Yup, it was right, and relativity give a large number, neat...

Answer 45

Cool! What was the order of the large number?

Answer 46

\(10^?\)

Answer 47

Not by much actually like I was using different numbers as this one was straight from the book but classically I got around 4m and with relativity 14m, still neat I guess, I'm assuming relativity is more accurate?