How Does the Positive Spring Constant k = F/x Apply in Bow Mechanics?

  • #1
Ebby
41
14
Homework Statement
What is the speed of the arrow when the bow string reaches its equilibrium position?
Relevant Equations
F = -kx and F = kx
U = 0.5kx^2
K.E. = 0.5mv^2
bow.JPG

The question is easy. I merely have a query:

When the bow string is released, the potential energy stored in it ##U = \frac {kx^2} {2}## is all transformed to kinetic energy ##K = \frac {mv^2} {2}##, so we have:$$v = \sqrt {\frac {k} {m}}x$$
I now need to eliminate ##k##, so I can use ##k = -F/x##:$$v = \sqrt {\frac {\frac {-F} {x}} {m}}x$$
Obviously this is no good because I need to take the square root of a negative quantity. So I use ##k = F/x## instead:$$v = \sqrt {\frac {\frac {F} {x}} {m}}x$$
My question is how using ##k = F/x## (i.e. the postive version of this equation) logically follows in the maths. Is it because the potential energy was originally put into the bow string by me doing positive work on the string, where the force is in the direction of the displacement. Or put another way, I draw the string back, and hence ##k = F/x## applies instead of ##k = -F/x##?
 
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  • #2
Ebby said:
I draw the string back, and hence k=F/x applies instead of k=−F/x?
Yes. ##F=-kx## is for where F is the force exerted by the spring. For the work done on the spring you need the force applied to the spring, which is -F.
 
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Likes Ebby
  • #3
Thank you :)
 
  • #4
Whether you pull on the stretched string or push on it the work that the string does on your finger is negative because the displacement is opposite to the force. That's what the minus sign in ##F=-kx## signifies: For positive ##x## you have negative ##F## and vice-versa. Thus, the work done by the string on your finger as the string is displaced from point A to point B is ##W_{AB}=-\frac{1}{2}kx_{AB}^2.##

Now the change in potential energy stored in the string is, by definition, the negative of the work done by the string. Two negatives make a positive, therefore the change in potential energy is always ##\Delta U_{AB}=+\frac{1}{2}kx_{AB}^2.##

The error in your solution is here:
Ebby said:
I now need to eliminate k, so I can use ##k=−F/x##
The constant ##k## cannot be a negative quantity because the amount of Newtons per meter that the string exerts when displaced is the same regardless of direction. The correct way to write it is ##k=|F/x|.##

Thus, your answer should have been $$v = \sqrt {\frac{|F/x|}{m} }|x|.$$ The absolute sign outside the radical is there because you are required to provide the speed which is always positive.
 
  • #5
When you use F=-kx it is assumed that the quantities "F" and "x" are signed quantities and not just manitudes. The minus in the formula shows that the force exerted by the spring is alwyas in the opposite direction to the displacement "x".

Which one of the two is positive and which is negative is a matter of choice, but they are in opposite directions so they have opposite signs. For example, if you pull the string to the right (to stretch it), the displacement is to the right and the force exerted by the spring is to the left. If you pick positive direction to be rightwards, then x is positive and the force is negative. If you pick left as the positive direction, then the force is positive and the displacement negative. For both choices k results to be positive.

The bottom line, if you use the formula with sign (minus in front of "kx") either the force or the displacement is negative (it does not matter which is which) and the constant k is positive. If you take both quantities as positive then you assume just the relationshop between magnitudes (F=kx) without a minus sign.
 

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