Question

How much elastic energy is stored in a slingshot that has a spring constant of 300 N/m and is stretched by 0.45m?

Answer 1

Answer:

E = 30.37 J

Step-by-step explanation:

Given that,

The spring constant of the spring, k = 300 N/m

The spring is stretched by 0.45 m, x = 0.45 m

We need to find the elastic energy stored in the spring. The formula for the elastic energy is given by :

$E=\dfrac{1}{2}kx^2\\\\=\dfrac{1}{2}\times 300\times (0.45)^2\\\\=30.37\ J$

So, the required elastic energy is equal to 30.37 J.