Determine the speed of a rock shot with a rubber band

A 47 g rock is placed in a slingshot and the rubber band is stretched. The force of the rubber band on the rock is shown by the graph in the figure (See the link below).

http://vvcap.net/db/tEopQRBXDgPCppPItKmQ…

The rubber band is stretched 27 cm and then released. What is the speed of the rock?

I found the spring constant (k) to be 200 N/m by using the graph (note that the distances are in cm and not m)

I can calculate the spring force by simply multiplying the spring constant by the displacement of .27 m to get a spring force of 54 N.

I am lost though how to get the velocity here... Help?

Ah, to get the launch speed, you need to know how much energy, PE, was stored in the sling shot. That stored energy will be converted into kinetic energy KE = 1/2 mU^2 = 1/2 k dX^2 = PE when the slingshot is released from dX = .27 m extension. k = 200 N/m if you are correct.

So U = sqrt(k/m) dX = sqrt(200/.047)*.27 = ? mps, where m = .047 kg is the rock mass. You can do the math.

So you don't use the force you found. You use the conservation of energy and the stored energy of the slingshot when pulled back.