Find the mechanical energy of a block-spring system having a spring constant of 0.93 N/cm and an oscillation amplitude of 4.2 cm.
total energy= 1/2 k (max amplitude)^2
To find the mechanical energy of a block-spring system, we need to consider the energy due to both the kinetic energy and the potential energy of the system.
The potential energy of a spring is given by the formula: PE = (1/2)kx², where k is the spring constant and x is the displacement from the equilibrium position.
In this case, the spring constant is given as 0.93 N/cm. However, it is useful to convert it to SI units, which is N/m. To convert, divide the given value by 100: 0.93 N/cm = 0.0093 N/m.
The oscillation amplitude is given as 4.2 cm. Again, it is useful to convert it to SI units, which is meters. To convert, divide the given value by 100: 4.2 cm = 0.042 m.
Now, we can find the potential energy of the system using the formula:
PE = (1/2)kx²
= (1/2)(0.0093 N/m)(0.042 m)²
Simplifying this expression, we get:
PE ≈ 1.22 x 10^-5 J
This is the potential energy of the block-spring system.
To find the total mechanical energy, we need to consider the contribution of kinetic energy. At the maximum displacement, all the potential energy is converted into kinetic energy. Therefore, the mechanical energy of the system is equal to the potential energy at maximum displacement:
ME = PE ≈ 1.22 x 10^-5 J
So, the mechanical energy of the block-spring system is approximately 1.22 x 10^-5 Joules.