Question

A 84g arrow is fired from a bow whose string exerts an average force of 125N on the arrow over a distance of 81cm .What is the speed of the arrow as it leaves the bow?
I converted 84g to .084kg, and 81cm to .81m. I divided (125N/.81m) to get 154 N/m
I used the formula: K1+UG1+US1-MewFn=K2+UG2+US2. I crossed out K1 since there is no kinetic energy before the arrow is fired. I crossed out Ug1, -MewFn, UG2, and US2 which only leaves me with US1=K2:
(1/2)kx^2=(1/2)mv^2....(the 1/2's cancel out)
Square root (kx^2/m)=v
Square root(154N/m * .81m/.084kg)

Can anyone help me with my mistake?

Answer 1

The amount of work done on the arrow is Fd, not F/d.

See the problem in your work above?

Answer 2

In the beginning where I divided (125N/.81m) Would I multiply them instead and get around 101Nm?? And then I would plug it in "Square root(101Nm * .81m/.084kg)."

Answer 3

The work done on the arrow by the bow is \( W = Fd\). Now, where does all the energy of that work go? Into changing the kinetic energy, KE, of the arrow.

If the arrow starts from rest, its initial KE is zero. If the arrow at the end of being worked on has velocity \( v \), then its KE is \( \frac{1}{2}mv^2 \) where \( m \) is the mass of the arrow.

Therefore
\[ Fd = \Delta KE = \frac{1}{2}mv^2 \]
Now solve for \( v \).