Anonymous
Anonymous asked in Science & MathematicsPhysics · 7 months ago

# Physics lab?

Help needed, I've read all the material but I still don't know how to answer these questions.

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• NCS
Lv 7
7 months ago

The second graph looks wonky, and there is a lot you haven't told us. For instance: is the length along the bottom the total length of the spring, or just the amount of stretch?

Looking at the first graph, the slope F / x gives the spring constant k. The bottom corner appears to be (x, F) = (0.45m, 1.3N) and the top right corner appears to be (x, F) = (1.18m, 4.9N). Then

k = (4.9-1.3)N / (1.18-0.45)m = 4.93 N/m

From this we can find the length of the elastic band (which I think is important):

F = kΔx

and using the top right point again,

4.9 N = 4.93N/m * Δx

Δx = 0.994 m

which suggests to me that the "spring" is

1.18m - 0.99m = 0.19 m long

So, for that point,

Eelastic = ½k(Δx)² = ½*4.93N/m*(0.99m)² = 2.435 J

and you are supposed to "prove" that this is equal to the change in GPE -- m*g*Δx.

But you haven't given us the mass, either.

Hopefully, this will get you moving along the path for solution. If so, please award Best Answer!