Prescribed practical 2 - method and conclusion
Apparatus
A steel spring, a 100g mass hanger, 12 x 100g masses, a retort stand, a boss and clamp, a clamp, a metre rule, an s-hook, a pointer, safety goggles, a slotted base.
Method
- Set up apparatus as shown in the diagram. Use a slotted base to secure the metre stick and ensure that it is vertical.
- Attach the mass hanger s -hook and pointer to the lower end of the spring. The pointer should just touch the metre rule.
- Read the pointer value from the metre rule. Record this length in a suitable table. This is the initial length of the spring for zero mass. We can neglect the mass of the hanger.
- Add a 100g slotted mass to the hanger. Record the mass in kg in the table.
- Read the new position of the pointer on metre rule. This is the stretched length of the spring. Record this length in the table.
- Calculate the stretching force = weight of masses: W = mg.
- Calculate: extension = stretched length 鈥 original length.
- Repeat the procedure by adding 100g masses in steps of 100g up to 1200g. Record the new stretched length each time by reading the position of the pointer on the metre rule. Subtract the original length from the new stretched length to calculate each extension.
Error
The main cause of error in this experiment is reading the stretched length of the spring.
The metre rule scale should be read at eye level directly opposite the pointer.
Use the slotted base to ensure that the metre rule is vertical.
Results
Initial length of the spring = X cm.
Mass in kg | Stretching force (F in N) | Stretched length in cm | Extension e in cm |
0.1 | |||
0.2 | |||
0.3 | |||
0.4 | |||
0.5 | |||
0.6 | |||
0.7 | |||
0.8 | |||
0.9 | |||
1.0 | |||
1.1 | |||
1.2 |
Mass in kg | 0.1 |
---|---|
Stretching force (F in N) | |
Stretched length in cm | |
Extension e in cm |
Mass in kg | 0.2 |
---|---|
Stretching force (F in N) | |
Stretched length in cm | |
Extension e in cm |
Mass in kg | 0.3 |
---|---|
Stretching force (F in N) | |
Stretched length in cm | |
Extension e in cm |
Mass in kg | 0.4 |
---|---|
Stretching force (F in N) | |
Stretched length in cm | |
Extension e in cm |
Mass in kg | 0.5 |
---|---|
Stretching force (F in N) | |
Stretched length in cm | |
Extension e in cm |
Mass in kg | 0.6 |
---|---|
Stretching force (F in N) | |
Stretched length in cm | |
Extension e in cm |
Mass in kg | 0.7 |
---|---|
Stretching force (F in N) | |
Stretched length in cm | |
Extension e in cm |
Mass in kg | 0.8 |
---|---|
Stretching force (F in N) | |
Stretched length in cm | |
Extension e in cm |
Mass in kg | 0.9 |
---|---|
Stretching force (F in N) | |
Stretched length in cm | |
Extension e in cm |
Mass in kg | 1.0 |
---|---|
Stretching force (F in N) | |
Stretched length in cm | |
Extension e in cm |
Mass in kg | 1.1 |
---|---|
Stretching force (F in N) | |
Stretched length in cm | |
Extension e in cm |
Mass in kg | 1.2 |
---|---|
Stretching force (F in N) | |
Stretched length in cm | |
Extension e in cm |
Graph
Plot a graph of stretching force, F in N on the y-axis, against extension, e in cm on the x-axis.
Join the points with a line of best fit.
Conclusion
We can see from the graph that as the stretching force increases the extension of the spring also increases. This agrees with our prediction.
In fact, since the line of best fit is a straight line through the origin, up to a certain point, we can be even more precise.
We can say that the stretching force F is directly proportional to the extension e up to a limit known as the limit of proportionality.
This is known as Hooke鈥檚 law.
Equation
Stretching force F=spring constant k x extension e
F = ke
The gradient of the graph = \(\frac {stretching~force~F}{extension~e}\) = spring constant k
Calculate the gradientIn a graph, the gradient is the steepness of the line. The greater the gradient, the greater the rate of change. of the line.