f=ma,

so assuming constant force while accellerating:

a=f/m

v=(f/m)t

s=(f/m) (t^2/2).

So assume a 1m pull on the bow, we have the time on the bow is:

t=sqrt(2m/f)

and so the speed at which the arrow leaves the bow is:

v=sqrt(2f/m).

Hence doubling the force increases the speed by a factor of sqrt(2).

Alternativly, work done = change in KE = force*distance.

Hence doubling the force doubles the KE, but KE is proportional to the

square of speed, so speed incread=ses by a factor of sqrt(2).

RonL