spring constant question

[evoviiigsr]

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Lets say that you have a horizontal spring that extends out to a pulley, where a mass M hangs onto the spring. 4 different measurements of the total change in spring length (L) are taken for 4 different masses(m).

trial 1: m=0 kg L=0.400
trial 2: m=0.1 kg L=0.405
trial 3: m=0.4 kg L=0.420
trial 4: m=2.0 kg L=0.500

Find the spring constant. The answer is 200 N/m, but I can't figure out why.

 

[emquiksilver]

F = ma for the object equals F = -kx for the spring. Set them equal to each other. x equals the change in length, so don't use the given values of L. L when m = 0 is the default position.

 

[RoadRunner17]

Just like the previous poster said, set forces equal to each other, in this case mg = kx where x is the difference between L and its equilibrium length Lo. So,

k = mg/(L-Lo) where Lo = 0.400m.

 

[Creightonite]

Lets say that you have a horizontal spring that extends out to a pulley, where a mass M hangs onto the spring. 4 different measurements of the total change in spring length (L) are taken for 4 different masses(m).

trial 1: m=0 kg L=0.400
trial 2: m=0.1 kg L=0.405
trial 3: m=0.4 kg L=0.420
trial 4: m=2.0 kg L=0.500

Find the spring constant. The answer is 200 N/m, but I can't figure out why.

trial #4 calculation:

equilibrium position: 0.4m (from trial #1)

so F=-k*deltaX

2*10=k*(0.5-0.4)

20=0.1k

k=200

 

[evoviiigsr]

Thanks guys. I'm officially an idiot