Calculating k in Hooke's Law: Understanding Archery and Projectile Motion

[aatgomez]

In preparing to shoot an arrow, an archer pulls a bowstring back 20 cm=.2 m. The arrow can reach a horizontal distance of 300 m and the arrow weighs 200 g. Find k.

m=.2 kg
W=1/2kd final- 1/2kd initial
Fs=-kx

Not really sure where to begin, and I've been trying for a while now.

 

[mfb]

There are squares missing in your second equation.

To shoot 300m, how fast does the arrow have to be initially? And what is the ideal angle to shoot the arrow, neglecting air resistance?

 

[aatgomez]

oh yes, sorry W=1/2kd^2 final - 1/2kd^2 initial
It would have to be 0 initially and I'm thinking 45 degrees?

 

[mfb]

45 degrees is right. With "initially", I meant the flight path, so directly after it leaves the bow. If it is too slow, it won't make 300 meters, obviously.

 

[aatgomez]

Ok, so if I find the initial velocity and the final velocity as if I were doing projectile problem then I can find W=1/2mvf^2 - 1/2mvi^2 then I could set W equal to 1/2kd^2 final - 1/2kd^2 initial and find k. Is that correct?

 

[mfb]

No. The energy from the spring does not get "used up" in the flight. It is still in the arrow when it hits the ground.

The velocity is the key connection here.

 

[aatgomez]

I see that there's potential energy before the arrow is shot. That's where the .2 m come in. But I still don't know how to solve it. A little more help please?

 

[mfb]

Find the velocity the arrow needs to fly 300m. Once you have that, the rest should follow from that velocity. Note that is the same hint I gave in post #2.

 

[aatgomez]

Right I realize that, that's what I don't know how to find.

 

[mfb]

The arrow is in free fall. Maybe the opposite problem is easier: If you know the initial velocity and angle, how far will it go?
I'm quite sure you have solved a problem of this type before in your course or as homework.