Calculating Arrow Speed with Doubled Force | Newton's Law Problem

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Homework Help Overview

The problem involves calculating the speed of an arrow when the force exerted by the bow is doubled, starting from an initial speed of 29.5 m/s. The context is rooted in Newton's laws of motion and the relationship between force, mass, and acceleration.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between force and acceleration, with one attempting to directly multiply the initial speed by two. Another participant calculates the new acceleration based on the doubled force and questions the assumption of constant time.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Some participants have provided guidance on the relationship between force and acceleration, while others have pointed out potential errors in reasoning regarding initial conditions and time.

Contextual Notes

There are assumptions about the constancy of distance and the nature of the bow being used, which may affect the calculations. The original poster's initial conditions and approach are also under scrutiny.

ethrust2
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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..
 
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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
 
thanks so much man
 
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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