Question

A spring-loaded toy gun is used to launch a 12.0-g plastic ball. The spring which has a spring constant of 15.0 N/m is compressed 8.00 cm as the ball is pushed into the barrell. When the trigger is pulled the swing is released and shoots the ball back out horizontally. What is the elastic potential energy stored in the spring when it is compressed 8.00 cm?

Answer 1

Answer:

the elastic potencial energy stored when the spring is compressed x=8 cm is K(x)= 0.048 J

Explanation:

since the work is related with the force through

W=∫F dx

for a spring of constant k :

F=k*x , where F= compression force, x= compression length

then for a compression from 0 until x

W=∫F dx = ∫ k*x dx =  k ∫x dx =1/2*k*x² - 1/2*k*0² = V(x) - V(0)

since the work depends only on the final value of compression and not on the process 1/2*k*x² represents the elastic potential energy V stored in the spring, then

V(0) = 1/2*k*0² = 0  

V(x) = 1/2*k*x²

when the spring is compressed x= 8 cm = 0.08m , the elastic potencial energy is

V (x) = 1/2*k*x² = 1/2 * 15 N/m* (0.08m)² = 0.048 J