A mass is connected to a spring and is hanging down at rest as shown in the figu
ID: 1287444 • Letter: A
Question
A mass is connected to a spring and is hanging down at rest as shown in the figure at the right. The dotted line indicated where the spring would be at rest if no mass were connected to it. For the period of time we are considering (2.5 s), internal damping of the spring and air resistance can be ignored. We consider two cases: A- We pull the mass down by 10 cm and release it. B- We pull the mass down by 30 cm and release it. In the figures below are shown graphs of the y-position vs. time using the ruler as scale (but in unit of meters rather than cm), labeled "A" and "B" for our two cases. The eight graphs below were calculated on a spreadsheet and represent the gravitational potential energy, the elastic potential energy, the kinetic energy, and the total energy for each of the two cases. The units are Joules. Fill out the table by matching the graphs below with each of these quantities in these cases A and B. If none work, put 9.Explanation / Answer
when mass is at its lowest position PE is maximum, Gravitational PE is min and KE is zero
when position is in mid point KE is maximum
at top most position PE is min GPE is max KE=0
GPE follows same pattern as Y vs t as GPE= mgh => option 3 for case B and option 4 for case A
Elastic PE = KX2/2 where X is change in length of spring, at mid point EPE = 0 and should incerase in either sides of mid position both A, B have mid positions at Y = 0.48, EPE graph should have its minima at 0.5 and maxima of B larger than Maxima of A optin 5 and 6 seems similar 5 for A and 6 for B, as Y starts from top most point EPE graph should start from 0
KE graph should start from 0 to max at 0.48 and to 0 again so none of these options are correct
Total energy is always constant = KE + PE and is larger for larger displacement hence 7 for A and 2 for B
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