Solve Spring Constant Homework Confusion

[physicaled]

Homework Statement

The question is "Imagine you are told after performing your experiment that the weights in your weight set are incorrectly labeled, and weigh less than you thought. Does that mean the value you determined for the spring constant is wrong? If so, is your value too high or too low?"

The variables are: the distanced stretched by the spring (dependent) and the weight used (in grams- independent)

Some data:

• We are to use the line of best fit to find the spring constant and two points on the line of best fit are: (5 g, 11.5 cm) and (45 g, 19 cm)
• The line of best fit was calculated to be 7.5 cm/40g or .1875 cm/g
  
• Other data points from the experiment are: (5 g, 11.75 cm), (10 g, 12.5 cm), (15 g, 13.5 cm) and (20 g, 14.5 cm)

Homework Equations

• Technically we are not to used the spring constant equation as we are trying to "figure it out" based upon this lab but F= -kx

The Attempt at a Solution

[/B]
This is where the confusion sets in because I see it in two different ways:

1. The spring constant found on the data would be too high because now, assuming the lengths measured have not changed, it can be said that it takes less weight to go more distance resulting in a lower spring constant
2. OR: If you assume the weights were 1/5 of what you thought they were two points on the new line of best fit could be (1 g, 11.5 cm) and (8 g, 19 cm) (these were taken from points on the original line of best fit). Then, when dividing you would be dividing by a smaller number resulting in a higher spring constant than originally calculated.

So I'm not really sure which is correct?

 

[Chestermiller]

Homework Statement

The question is "Imagine you are told after performing your experiment that the weights in your weight set are incorrectly labeled, and weigh less than you thought. Does that mean the value you determined for the spring constant is wrong? If so, is your value too high or too low?"

The variables are: the distanced stretched by the spring (dependent) and the weight used (in grams- independent)

Some data:

• We are to use the line of best fit to find the spring constant and two points on the line of best fit are: (5 g, 11.5 cm) and (45 g, 19 cm)
• The line of best fit was calculated to be 7.5 cm/40g or .1875 cm/g
  
• Other data points from the experiment are: (5 g, 11.75 cm), (10 g, 12.5 cm), (15 g, 13.5 cm) and (20 g, 14.5 cm)

Homework Equations

• Technically we are not to used the spring constant equation as we are trying to "figure it out" based upon this lab but F= -kx

The Attempt at a Solution

[/B]
This is where the confusion sets in because I see it in two different ways:

1. The spring constant found on the data would be too high because now, assuming the lengths measured have not changed, it can be said that it takes less weight to go more distance resulting in a lower spring constant
2. OR: If you assume the weights were 1/5 of what you thought they were two points on the new line of best fit could be (1 g, 11.5 cm) and (8 g, 19 cm) (these were taken from points on the original line of best fit). Then, when dividing you would be dividing by a smaller number resulting in a higher spring constant than originally calculated.

So I'm not really sure which is correct?

To get the spring constant, you divide the force change by the length change, not the other way around.

 

[physicaled]

force change by the length change

To get the spring constant, you divide the force change by the length change, not the other way around.

So if it is then 5.33 g/cm for the constant then when the weights are said to be less, the spring constant would decrease? Because 40 g/7.5 cm becoming 7g/7.5 cm would be a whole lot less.

 

[Chestermiller]

force change by the length changeSo if it is then 5.33 g/cm for the constant then when the weights are said to be less, the spring constant would decrease? Because 40 g/7.5 cm becoming 7g/7.5 cm would be a whole lot less.

Yes. $$k=\frac{\Delta F}{\Delta x}$$