A) k = m g / x = 5.0 * 9.8 / .20 N/m
B) find the work done (energy) in stretching the spring from .20 m to .80 m
... 1/2 k x^2 ... for each position
the work is energy stored in the spring
... when the mass is released, the stored energy accelerates it upward
... the stored energy, minus the change in position (gravitational energy)
is the kinetic energy of the mass at the equilibrium point
A block with mass of 5.0kg is suspended from an ideal spring having negligible mass and stretches the spring 0.20m to its equilibrium position.
A) What is the force constant of the spring?
B) The spring is then stretched 0.80m from its equilibrium position and then released with the velocity zero. Calculate the velocity of the mass when the mass passes again through the equilibrium position.
1 answer
1 answer