Question
mnol=50g
xnol=19cm
M=(m+mnol) plus or minus 0.05cm 250, 450, 650,850,1050
loading =25,30,35,40,45,45
unloading =25,30,35,40,45
Mean,pointer (x plus or minus 0.05cm)= 25,30,35,40,45
Extension e=(x-xnol) plus or minus 0.05cm 6cm , 11cm, 16cm, 2cm, 26cm
1) plot a graph of M against e
2) measure the slope and explain its physical meaning. state its unit and estimate the error
3)did your graph pass through the origin? if not, give reasons
4) deduce from your load ( mass) extension, graph the weight Wa of the given object, A .
5) estimate the error in Wa
xnol=19cm
M=(m+mnol) plus or minus 0.05cm 250, 450, 650,850,1050
loading =25,30,35,40,45,45
unloading =25,30,35,40,45
Mean,pointer (x plus or minus 0.05cm)= 25,30,35,40,45
Extension e=(x-xnol) plus or minus 0.05cm 6cm , 11cm, 16cm, 2cm, 26cm
1) plot a graph of M against e
2) measure the slope and explain its physical meaning. state its unit and estimate the error
3)did your graph pass through the origin? if not, give reasons
4) deduce from your load ( mass) extension, graph the weight Wa of the given object, A .
5) estimate the error in Wa
Answers
GPT-4o mini
To analyze your given data and complete the tasks, I will break down the necessary steps in detail.
### 1) Plot a Graph of M Against e
First, we need to identify the values for \( M \) (mass plus the mass of the nominal weight \( mnol \)) and \( e \) (extension).
Given:
- \( mnol = 50g = 0.05 \, kg \)
- \( M \) values (in kg):
- For 250g: \( M = 0.25 + 0.05 = 0.30 \, \text{kg} \)
- For 450g: \( M = 0.45 + 0.05 = 0.50 \, \text{kg} \)
- For 650g: \( M = 0.65 + 0.05 = 0.70 \, \text{kg} \)
- For 850g: \( M = 0.85 + 0.05 = 0.90 \, \text{kg} \)
- For 1050g: \( M = 1.05 + 0.05 = 1.10 \, \text{kg} \)
The corresponding extension \( e \) values (in cm) are:
- 25g: 6 cm
- 30g: 11 cm
- 35g: 16 cm
- 40g: 21 cm
- 45g: 26 cm
You can now create a scatter plot, with \( M \) on the x-axis and \( e \) on the y-axis.
### Data for Plotting
| Mass (M) | Extension (e) |
|----------|---------------|
| 0.30 kg | 6 cm |
| 0.50 kg | 11 cm |
| 0.70 kg | 16 cm |
| 0.90 kg | 21 cm |
| 1.10 kg | 26 cm |
### 2) Measure the Slope and Explain Its Physical Meaning
The slope of the line in your graph will represent the stiffness (spring constant \( k \)) of the material according to Hooke's Law \( F = k \cdot e \).
Unit:
- The unit for slope is \( \text{kg/cm} \) (or \(\text{N/m}\) if you convert cm to m).
To find the slope:
- Use two well-defined points from your line, for example, (0.30, 6) and (1.10, 26).
Slope \( k \) = \( \frac{\Delta e}{\Delta M} \)
Error estimation for the slope can depend on how accurately you place your line of best fit and the uncertainty in measurements.
### 3) Did Your Graph Pass Through the Origin?
The graph may not pass through the origin because:
1. There might be an initial load applied to the spring before any extension occurs (preload).
2. Laboratory equipment might not have a zero point that aligns perfectly.
3. The system could exhibit elastic behavior only after surpassing a small initial tension.
If it doesn’t pass through the origin, record the y-intercept value and discuss.
### 4) Deduce from Your Load (Mass) Extension, the Weight \( W_a \) of the Given Object A
Using the weight formula:
\[
W = m \cdot g
\]
Where \( g \approx 9.81 \, \text{m/s}^2 \).
You will want to calculate the weight \( W_a \) for the given masses:
- For 250g: \( W_a = 0.25 kg \cdot 9.81 = 2.45 N \)
- For 450g: \( W_a = 0.45 kg \cdot 9.81 = 4.41 N \)
- Continue this for all calculated masses.
### 5) Estimate the Error in \( W_a \)
The error in \( W_a \) can be approximated based on the errors in mass measurement and considering uncertainties in \( g \).
You’ll need to consider:
- The error in mass \( \Delta m = 0.05 \, g \)
- The error propagation due to gravity \( \Delta W_a = g \cdot \Delta m \)
To put this together accurately, you can derive the total error using the formula, keeping in mind the individual uncertain factors you accounted for.
### Final Steps
Make sure to visually depict the graph, calculate the slope carefully, and confirm your findings through the physics principles governing the elasticity of materials. If you have additional software (like Excel) or graphing tools, they can greatly help with creating clear visuals and conducting calculations quickly.
### 1) Plot a Graph of M Against e
First, we need to identify the values for \( M \) (mass plus the mass of the nominal weight \( mnol \)) and \( e \) (extension).
Given:
- \( mnol = 50g = 0.05 \, kg \)
- \( M \) values (in kg):
- For 250g: \( M = 0.25 + 0.05 = 0.30 \, \text{kg} \)
- For 450g: \( M = 0.45 + 0.05 = 0.50 \, \text{kg} \)
- For 650g: \( M = 0.65 + 0.05 = 0.70 \, \text{kg} \)
- For 850g: \( M = 0.85 + 0.05 = 0.90 \, \text{kg} \)
- For 1050g: \( M = 1.05 + 0.05 = 1.10 \, \text{kg} \)
The corresponding extension \( e \) values (in cm) are:
- 25g: 6 cm
- 30g: 11 cm
- 35g: 16 cm
- 40g: 21 cm
- 45g: 26 cm
You can now create a scatter plot, with \( M \) on the x-axis and \( e \) on the y-axis.
### Data for Plotting
| Mass (M) | Extension (e) |
|----------|---------------|
| 0.30 kg | 6 cm |
| 0.50 kg | 11 cm |
| 0.70 kg | 16 cm |
| 0.90 kg | 21 cm |
| 1.10 kg | 26 cm |
### 2) Measure the Slope and Explain Its Physical Meaning
The slope of the line in your graph will represent the stiffness (spring constant \( k \)) of the material according to Hooke's Law \( F = k \cdot e \).
Unit:
- The unit for slope is \( \text{kg/cm} \) (or \(\text{N/m}\) if you convert cm to m).
To find the slope:
- Use two well-defined points from your line, for example, (0.30, 6) and (1.10, 26).
Slope \( k \) = \( \frac{\Delta e}{\Delta M} \)
Error estimation for the slope can depend on how accurately you place your line of best fit and the uncertainty in measurements.
### 3) Did Your Graph Pass Through the Origin?
The graph may not pass through the origin because:
1. There might be an initial load applied to the spring before any extension occurs (preload).
2. Laboratory equipment might not have a zero point that aligns perfectly.
3. The system could exhibit elastic behavior only after surpassing a small initial tension.
If it doesn’t pass through the origin, record the y-intercept value and discuss.
### 4) Deduce from Your Load (Mass) Extension, the Weight \( W_a \) of the Given Object A
Using the weight formula:
\[
W = m \cdot g
\]
Where \( g \approx 9.81 \, \text{m/s}^2 \).
You will want to calculate the weight \( W_a \) for the given masses:
- For 250g: \( W_a = 0.25 kg \cdot 9.81 = 2.45 N \)
- For 450g: \( W_a = 0.45 kg \cdot 9.81 = 4.41 N \)
- Continue this for all calculated masses.
### 5) Estimate the Error in \( W_a \)
The error in \( W_a \) can be approximated based on the errors in mass measurement and considering uncertainties in \( g \).
You’ll need to consider:
- The error in mass \( \Delta m = 0.05 \, g \)
- The error propagation due to gravity \( \Delta W_a = g \cdot \Delta m \)
To put this together accurately, you can derive the total error using the formula, keeping in mind the individual uncertain factors you accounted for.
### Final Steps
Make sure to visually depict the graph, calculate the slope carefully, and confirm your findings through the physics principles governing the elasticity of materials. If you have additional software (like Excel) or graphing tools, they can greatly help with creating clear visuals and conducting calculations quickly.