Simple newton law problem

  • Thread starter ethrust2
  • Start date
  • #1
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Homework Statement


An arrow, starting from rest, leaves the bow with a speed of 29.5 m/s. If the average force exerted on the arrow by the bow were doubled, all else remaining the same, with what speed would the arrow leave the bow?


Homework Equations



use a= v-vo/t and f=ma

The Attempt at a Solution



i tried to just multiply the speed by 2 but that's wrong... and the same speed is wrong too..
 

Answers and Replies

  • #2
4
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F = ma
If the force doubled, then the acceleration must double, because the mass definitely can't double xD

So, that means the new acceleration is twice the initial acceleration. This doesn't mean the new velocity is twice the old velocity.

First, let's find the initial acceleration.
vf = vi + at
29.5 = 0 + at
ai = 59/t

The new acceleration is twice the initial acceleration, so
anew = (2*59)/t = 118/t
anew = = 118/t

Now, plugging that back into the equation:
vf = vi + at
vf = 29.5 + (118/t)*t ===> Note how the t cancels out
vf = 29.5 + 118
vf = 147.5 m/s
 
  • #3
3
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thanks so much man
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

Hi Firecloak! Welcome to PF! :smile:

Sorry, but your method is completely wrong …

vi is still zero, not 29.5; and it isn't the same t. :redface:

Hi ethrust2! :wink:

The distance will be the same (exactly the same if it's a crossbow, and almost exactly if it's a longbow) …

so … using work done and conservation of energy … how does v depend on F? :smile:
 

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