A 200-g mass is attached to a spring as in DL. The mass is lifted up 5 cm and re

Question

A 200-g mass is attached to a spring as in DL. The mass is lifted up 5 cm and released so that the mass begins to oscillate about the equilibrium point. The spring has a spring constant k = 500 N/m (J/m2). a) Calculate and accurately plot on a full size (8.5 x 11 in.) sheet of graph paper PEspring-mass, Etotal, and KE. The vertical axis of the graph should be energy (joules). The horizontal axis is “distance from equilibrium”. b) On the same graph, quickly sketch (without calculating values) the PE, KE and Etot of the system if the mass were initially pulled back (stretched) 2.5 cm from its equilibrium point, instead of lifted up (compressed) 5 cm

Explanation / Answer

Spring Potential Energy= 1/2kx^2 so 1/2(500)(25)= 6250 Joules.

This is also equal to the total Energy. Energy will transfer from SPE to Kinetic Energy as the mass oscillates. The mass will have a maximum Kinetic Energy of 6250 joules. As the mass approaches the equilibrium point it has no SPE and all of it is Kinetic.
So,
Initial
                      SPE=6250
                          here the   KE=0 , so the total energy
                         E=6250
Equilibrium Point
SPE=0
KE=6250
                  E=6250

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