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Hooke's Law. Does this spring...

by valeriex0x
Tags: hooke, spring
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valeriex0x
#1
Oct2-11, 09:21 AM
P: 43
1. The problem statement, all variables and given/known data

A student measures the force required to stretch a spring by various amounts and makes the graph shown in the figure, which plots this force as a function of the distance the spring has stretched.
Part A) Does this spring obey Hooke's Law?
Part B) What is the force constant of the spring in N/m?
Part C) What force would be needed to stretch the spring a distance of 17 cm from its unstretched length, assuming that it continues to obey Hooke's Law (F=N)?


2. Relevant equations

F is proportional to displacement x
F=-kx
k=delta f/delta x

3. The attempt at a solution

We didnt cover this in class yet, but I know that a hookean body gets displaced, and once the force is removed it goes back to its origional position. Which means that the displacement along the x axis is the same as the force applied to it.

For part A) I would say No, the spring does not obey Hooke's Law.
For part B) I get -1/2 N/m from F=-kx 5=(-x) 2.5
For part C) -17N from F=(-k ) 17cm
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EkaterinaAvd
#2
Oct2-11, 10:02 AM
P: 18
they might mean absolute values of force and displacement and you just should treat it as F=kx. Because negative k would mean that when your spring is stretched, it tends to be streched even more. And when it is contracted, it tends to be contracted even more.

(A) Why do you think so? And if it doesn't obey, how can you use it to solve parts (B) and (C)?

(B) Suppose you are right and k=-1/2. Then from your expression: F=-kx=-(-1/2 N/m)*2.5 m=1.25 N which is not equal to 5 N

(C) Suppose you were right in part (B) and k=-1/2. Then F=-(-1/2 N/cm)*17 cm = 8.5 N not equal to 17 N.

Please, think a little more about all three parts and post the next iteration of your solution.
valeriex0x
#3
Oct2-11, 10:36 AM
P: 43
I originally thought the answer to Part A) was No, because when I looked at the graph, when x=5, y is not equal to 5. I thought that is something obeyed Hooke's Law both the x and y value would be 5, and again when x=10, y would be 10. I entered the reponse "No" into mastering physics and they told me I was wrong, so obsivoulsy I entered Yes and then got it correct. Was I wrong in my thinking to believe that they are supposed to be equal? Or is it okay that the x and y values are not equal just so long as the graph is linear?

Can you just explain why my first answer was incorrect? Or did I kind of figure it out with the above response.

valeriex0x
#4
Oct2-11, 10:44 AM
P: 43
Hooke's Law. Does this spring...

For Part B I calculated:

ΔF/Δx
(15-7.5)/(10-5)=1.5

Am i on the right track?
EkaterinaAvd
#5
Oct2-11, 10:54 AM
P: 18
part (B) is correct now in N/cm. But if you need answer in SI, you need to transfer it to SI.

for part A,
you can't expect force and displacement to be equal to each other or not because they have different dimensions.

imagine that it had x=5 when y=5 etc. Now lets measure x in meters instead of centimeters. and you'd have x=0.05 when y=5 when nothing actually changed in your system.

Hooke's law just says that force depends on displacement linearly with some coefficient k, F=kx (talking about magnitudes). This coefficient compensates the difference in dimensions of F and x (N/cm or N/m).

As your graph is straight line, answer is "Yes", force depends on displacement linearly. If it were parabola or circle of sin or etc, you would answer "No".


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