Question

Hang the 100 g mass on spring 1 and enter the additional displacement of the spring

_____10______cm.   Calculate the spring constant of this spring___________ N/m.

3. Place the 250 g mass on spring 3 and change the “Softness” setting to 1 notch to the right of the middle.

Calculate the spring constant of spring 3 :_________________N/m.

Note that the displacement of the springs with the masses attached is the NEW equilibrium length of the spring. The restoring force of the spring is now equal to the weight of the mass.

4. A stretched or compressed spring contains elastic potential energy that is equal to the work (energy) that was originally used to stretch/compress it.
Elastic potential energy is PE=1/2 K*2 and has units of Joules.

Calculate the potential energy of a spring with k = 300.0 N/m, that has been stretched by 1.5 cm.

Potential Energy = ____________.___ Joules

8 3 2 2 8 50 cm 9

Answer 1

spring constant K = W/dy

W = weight on spring = m*g

m = mass attached to spring

dy = stretch in spring

spring constant = 0.1*9.8/0.1 = 9.8 N/m

===========================

stretch = 20 cm = 0.2 m

mass m = 250 g = 0.25 kg

spring constant = 0.25*9.8/0.2 = 12.25 N/m

======================

4)

potential energy U = (1/2)*k*x^2

U = (1/2)*300*0.015^2

U = 0.034 joules