Calculating Potential and Kinetic Energy in a Slingshot Physics Problem

[stanton]

Homework Statement

A slingshot consists of a light leather cup, containing a stone, that is pulled back against two parallel rubber bands. It takes a force of 15 N to stretch either one of these bands 1.0 cm.

(a) What is the potential energy stored in the two bands together when a 50-g stone is placed in the cup and pulled back 0.20 m from the equilibrium position?

(b) With what speed does the stone leave the slingshot?

The answers are a) PE = 60 J b) v = 49 m/s

Homework Equations

F=kx PE=1/2kx^2

The Attempt at a Solution

a)
F=kx
15 = k(0.01) → k=1500 N/m

PE = 1/2 k x^2
= 1/2 (1500)(.20)2
= 30 J

b)
PE1=KE1, KE = 1/2mv^2
1/2mv^2 = 30
1/2(0.050)v^2 = 30;
v= 34.6 m/s

Any comment or help would be appreciated about why I got the answers wrong. Thank you.

 

[stanton]

Ok, I realized what was wrong with the first question. Since there are two straps, 30 must be doubled.

 

[jegues]

Just remember, although you fixed your result for part A, you use it to answer part B. (In other words you have to make changes to part B as well)

 

[stanton]

Thank you so much. :)