A mess is attached to a spring. The rest position of the spring is 45 cm. The mass is then pulled to stretch the spring to a position of 51 cm. If the spring constant is 805 N/m. What is the elastic potential energy in the spring at this point?
We can use the formula for calculating the elastic potential energy in a spring:
Elastic Potential Energy = (1/2) * k * x^2
Where k is the spring constant and x is the displacement from the equilibrium position.
In this case, the spring constant is given as 805 N/m and the displacement from the equilibrium position is 51 cm - 45 cm = 6 cm = 0.06 m.
Plugging these values into the formula:
Elastic Potential Energy = (1/2) * 805 N/m * (0.06 m)^2
Elastic Potential Energy = 0.5 * 805 N/m * 0.0036 m^2
Elastic Potential Energy = 0.72576 N*m = 0.73 J
Therefore, the elastic potential energy in the spring at this point is 0.73 Joules.