Relationship between Velocity, Kinetic Energy and Mass

[Nile Anderson]

Homework Statement

Homework Equations

i) F=ma=mv/t
ii) E=mv²/2

The Attempt at a Solution

Now based on equation 1 , I have concluded that the velocity gained is inversely proportional to m, and that so is the kinetic energy , I have seen otherwise.[/B]

 

[BvU]

How can you conclude that if equation 1 does not contain any distance/displacement at alll ? You need other equations !

 

[Nile Anderson]

How can you conclude that if equation 1 does not contain any distance/displacement at alll ? You need other equations !

Oh hmmmm, x=at^2/2, which means, a=2x/t^2, which means that mv/t=2mx/t^2=F, so does this mean they are independent , I am sorry , I am a little weak.Could you please elaborate?

 

[BvU]

mv/t = F doesn't help you if you don't know how t depends on m.

Would the exercise be easier for you if the text would read:

You have a force $F$ to accelerate a mass $m$ from rest with constant acceleration over a distance $d$. What is the final velocity ?​

 

[Nile Anderson]

mv/t = F doesn't help you if you don't know how t depends on m.

Would the exercise be easier for you if the text would read:

You have a force $F$ to accelerate a mass $m$ from rest with constant acceleration over a distance $d$. What is the final velocity ?​

I think I see something , the energy gained by the system would be Fd=mad, so this implies that mv^2/2=mad, v^2/2=ad, so this would mean the kinetic energy is independent of m ? Further a=F/m , v=sqrroot(Fd/m), does that mean v varies in an inverse proportion with sqrt(m)

 

[Nile Anderson]

mv/t = F doesn't help you if you don't know how t depends on m.

Would the exercise be easier for you if the text would read:

You have a force $F$ to accelerate a mass $m$ from rest with constant acceleration over a distance $d$. What is the final velocity ?​

Sir ?

 

[BvU]

kinetic energy is independent of m

Correct

Further a=F/m , v=sqrroot(Fd/m), does that mean v varies in an inverse proportion with sqrt(m)

Bingo.

Can you also do the alternative route ? $\ d = {1\over 2} at^2\$ tells you $\ t \propto \sqrt d\$ and you already had $\ mv/t = F$

 

[Nile Anderson]

Correct
Bingo.

Can you also do the alternative route ? $\ d = {1\over 2} at^2} \$ tells you $\ t \propto \sqrt d\$ and you already had $\ mv/t = F$

Thank you so much sir , I get it now

 

[Nile Anderson]

Correct
Bingo.

Can you also do the alternative route ? $\ d = {1\over 2} at^2\$ tells you $\ t \propto \sqrt d\$ and you already had $\ mv/t = F$

Hmmmmm

 

[Nile Anderson]

Right , I get back the same place sir, thank you so much sir , that helped a lot.

 

[BvU]

Hmmm as in "delicious" or Hmmm as in "I don't believe a word of what I read" ?

 

[Nile Anderson]

Hmmm as in "delicious" or Hmmm as in "I don't believe a word of what I read" ?

lol "hmmm" as in intriguing

 

[BvU]

Right , I get back the same place sir, thank you so much sir , that helped a lot.

Ah, posts crossed. Well done. On to the next exercise.

PS

You have a force $F$ to accelerate a mass $m$ from rest with constant acceleration over a distance $d$. What is the final velocity ?​

was meant to seduce you to calculate v as a function of knowns:
$$\ d = {1\over 2} at^2\ \Rightarrow t = \sqrt{2d\over a} \Rightarrow v = at = {F\over m} \sqrt{2dm\over F} \Rightarrow v\propto 1/\sqrt m$$