What is the kinetic energy now?

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genu
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Homework Statement



Initially a body moves in one direction and has kinetic energy K. Then it moves in the opposite direction with three times its initial speed. What is the kinetic energy now?

Homework Equations



k =1/2mv^2

The Attempt at a Solution



k = 1/2m(3v)^2
k=1/2m(9v^2(
k=(m9v^2)/2
2k=m9v^2
2/9k=mv^2
...
?

answer: 9k
 
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You've actually found the answer within the first few steps.

As you've said, [tex]K = 1/2*mv^2[/tex]

And afterwards, [tex]K' = 1/2*m(-3v)^2[/tex]

so that [tex]K' = 9/2*mv^2[/tex]

What do you notice about K and K' when comparing them?

K' = 9K.
 
I'm not understanding how you've arrived to that last step...what happened to all the other variables?
 
genu said:
I'm not understanding how you've arrived to that last step...what happened to all the other variables?

You found that the final KE was 9 * (½mv²)

If you divide by the initial KE you get

Final KE = 9 * (Initial KE)

The suggestion was to encourage you to recognize the initial KE in the final KE expression.