a) Near the bottom of the spring, the spring only supports the mass of the mass
ID: 2065798 • Letter: A
Question
a) Near the bottom of the spring, the spring only supports the mass of the mass hanger (50 g). Assuming its k value is 12 N/m, how much does it want to stretch? (Distance) b) Now, near the top, the spring has the mass hanger (50 g) plus its own weight (75 g). Now how much does it want to stretch? c) Actually, the springs we had had a changing k; it was higher at the top than at the bottom. (That's why it looked tapered) Why it was that way? What do you think would have happened if you put it upsidedown?
Explanation / Answer
a.
F = kx
0.050*9.81 = 12*x
x = 0.0409m = 4.09cm
b.
F = kx
(0.050+0.075)*9.81 = 12*x
x = 0.101m = 10.1cm
c
I guess it looked that way because the part of the spring near the bottom was extending to its max before the force was evenly transfered to the upper part of the spring.
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