- #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

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##?