Force and Spring

Masses and Springs

1. Spring Constant

K1 = F/x1 = (.05kg)(9.8 N/kg) = 9.8 N/m .05m

K2 = F/x2 = (.1kg)(9.8 N/kg) = 9.8 N/m .1m

K3 = F/x3 = (.25kg)(9.8 N/kg) = 9.8 N/m .25m K avg. = 9.8 N/m

2. Unknown Masses
Green: m = kx = 9.8 * .07 = .07kg = 70g g 9.8

Gold: m = kx = 9.8 * .155 = .155kg = 155g g 9.8

Red: m = kx = 9.8 * .3 = .300kg = 300g 9.8

3. Explain the effect of mass on Hooke’s Law.
Hooke's Law states that the restoring force of a spring is directly proportional to a small displacement. Applying a mass to the end of the spring stretches it, changing the length from its original equilibrium position of rest, to a new (lower) equilibrium position. At this position, the vertical restoring force of the spring balances the weight and the downward pull of gravity is balanced by the upward pull of the spring. In simulation, by adding any of the mass to the spring, equilibrium position is changed, the more the mass the farther the spring stretches.
4. Explain the effect of gravity on Hooke’s Law by checking out what happens in the four different locations, as well as when there is no gravity.
Jupiter: The gravitational pull on Jupiter is 23.6N/kg, and this increases the weight of the 100g mass and the spring is stretched 26cm from equilibrium, making it a larger displacement than on any other planet.
Moon: The gravitational pull on the moon is 1.6N/kg, and this decreases the weight of the 100g mass. The spring is stretched 2cm. this is the smallest displacement compared to any other planet.
Earth: Under the weight of a 100g mass the spring is displaced 10cm from equilibrium due to gravitational pull of 9.8N/kg.
Planet X: The spring is displaced 4 cm from equilibrium because the weight of the100g mass decreases due to the amount of gravitational pull on the planet
Zero Gravity: At zero gravity the spring stays at equilibrium because there is no weight to the mass. In simulation the addition of any mass to the spring indicated no movement.
5. Explain the effect of the spring constant, k, (stiffness of spring) on Hooke’s Law.
The spring constant, or stiffness of the spring dictates how much stretch the spring will have away from equilibrium when a mass is attached to it. A stiff spring would have a high k value, and a soft spring would have a low k value. By adjusting the spring stiffness to a higher setting, less stretch is noted. Adjusting to a lower setting, more stretch of the spring is noted.
6. Explain the effect of friction on Hooke’s Law.
When a mass is fixed to a spring, it oscillates back and forth about a fixed position. Damping of the oscillation due to imposed or inadvertent resistive forces such as air resistance, internal resistance in the spring, or some externally applied mechanism causes the oscillations to eventually stop. In simulation, adding friction reduces oscillation, and by reducing friction oscillation is prevalent. This is indicated by continuous bouncing of the spring.