a block of wood of mass 0.25kg is attached to one end of a spirnf og constant stiffness 100N/m. The block can oscillate horizontally on a frictionless surface, the other end of the spring being fixed.
a) calculate the maximum elastic potential energy of the system for a horizontal oscillation of amplitude 0.20m.
i tried to use the equation
Ep=(1/2)*(mass)*(omega^2)*(displacement)
but i wasn't sure if this was the right way to go about it because i couldn't figure out how i could find out the angular frequency
thanks for your help!
Use: Ep = (1/2)kx^2 to get the elastic potential.
To calculate the maximum elastic potential energy of the system, you need to find the angular frequency first. The angular frequency is related to the spring constant and the mass of the block by the formula:
ω = sqrt(k / m)
where ω is the angular frequency, k is the spring constant, and m is the mass of the block.
Given that the spring constant k = 100 N/m and the mass m = 0.25 kg, you can plug these values into the formula to find the angular frequency:
ω = sqrt(100 N/m / 0.25 kg) = sqrt(400 N/kg) = 20 rad/s
Now that you have the angular frequency, you can use the formula you mentioned to calculate the maximum elastic potential energy:
Ep = (1/2) * m * ω^2 * displacement
Given that the displacement is 0.20 m, you can substitute the values:
Ep = (1/2) * 0.25 kg * (20 rad/s)^2 * 0.20 m
Ep = 0.5 J
Therefore, the maximum elastic potential energy of the system for a horizontal oscillation of amplitude 0.20 m is 0.5 Joules.