A spring with a spring constant k of 44 N/m is stretched a distance of 30 cm (0.3 m) from its original unstretched position. What is the increase in potential energy of the spring?
P.E. = (1/2) k X^2
Use X = 0.3 meters
The original value of X (deviation from the equilibrium position) was zero.
The answer will be in Joules
1.98
To determine the increase in potential energy of the spring, we can use the formula for potential energy stored in a spring:
Potential energy = (1/2) * k * x^2
where k is the spring constant and x is the displacement of the spring from its original unstretched position.
In this case, the spring constant is given as 44 N/m and the displacement is 0.3 m.
Substituting the given values into the formula, we have:
Potential energy = (1/2) * (44 N/m) * (0.3 m)^2
Calculating this expression:
Potential energy = (1/2) * 44 * 0.09
Potential energy = 1/2 * 3.96
Potential energy = 1.98 Joules
Therefore, the increase in potential energy of the spring is 1.98 Joules.