Suppose that a 200g mass (0.20kg) is oscillating at the end of a spring upon a horizontal surface that is essentially friction-free. The spring can be both stretched and compressed and have a spring constant of 240N/m. It was originally stretched a distance of 12cm (0.12m) from its equilibrium (un-stretched) position prior to release What is its initial potential energy?
PE=kA^2/2=240•(0.12)^2/2=1.728 J.
1.728J
To calculate the initial potential energy of the mass-spring system, we need to consider the spring constant and the displacement from the equilibrium position.
The potential energy of a mass-spring system is given by the equation:
Potential Energy (PE) = (1/2) * k * x^2
Where:
PE is the potential energy
k is the spring constant
x is the displacement from the equilibrium position
Given:
Mass (m) = 0.20 kg
Spring constant (k) = 240 N/m
Displacement (x) = 0.12 m
Substituting the given values into the equation:
PE = (1/2) * 240 N/m * (0.12 m)^2
Simplifying the equation:
PE = 0.5 * 240 N/m * 0.0144 m^2
Calculating:
PE = 20.736 Joules
Therefore, the initial potential energy of the mass-spring system is 20.736 Joules.
To calculate the initial potential energy of the mass-spring system, we need to consider the formula for potential energy stored in a spring.
The formula for potential energy in a spring is given by:
Potential Energy (U) = (1/2) k x^2
Where:
- U is the potential energy
- k is the spring constant
- x is the displacement from the equilibrium position
In this case, the spring constant (k) is given as 240 N/m. The displacement (x) from the equilibrium position is given as 0.12 m.
Now we can substitute these values into the formula and calculate the potential energy:
Potential Energy (U) = (1/2) * 240 N/m * (0.12 m)^2
Calculating this expression gives us:
U = (1/2) * 240 N/m * 0.0144 m^2
= 3.456 Joules
Therefore, the initial potential energy of the system is 3.456 Joules.