CSEC Physics Syllabus - Effective for examinations from May - June 2015
Section A - Mechanics
Statics
Deformation
Deformation
Specific Objective 3.13
investigate the relationship between extension and force;
When the load on the spring is 7.0 N, the length of the spring is 11.75 mm as seen from the graph above.
Use the information given in the diagram to calculate:
Specific Objective 3.14
solve problems using Hooke's law.
Pixabay
Example 1
The following results were obtained by a student when a spiral spring was stretched within its elastic limit.
| Load/N | 2.0 | 4.0 | 6.0 | 8.0 | 10.0 | 12.0 | 14.0 | 16.0 |
| Length/mm | 9.4 | 10.3 | 11.2 | 12.1 | 13.6 | 14.0 | 14.9 | 15.9 |
a) Plot a graph of Length against Load. (You can start the scale on the Length axis from a value of 8.0 mm.)
b) Which point on the graph looks as if an error might have been made by the student? You should ignore this point when drawing your 'best fit' line.
c) Find the gradient or the slope of the graph, S.
d) Find the spring constant. (The spring constant is the reciprocal of the slope, i.e. 1/S)
e) Use your graph to find the length of the spring when there is no load on the spring.
f) Find the length of the spring when the load is 7.0 N.
g) Calculate the extension of the spring for the same load.
h) What load would produce an extension of 4.0 mm?
i) How would the student know that in her experiment she had not taken the spring past its elastic limit?
Solution
a)
b) The error which might have been made by the student is at point (10.0 N, 13.6 mm), which is identified by the red circle.
c)
Gradient, S = (y2 - y1) ÷ (x2 - x1)
S = (15.5 mm - 9.1 mm) ÷ (15.2 N - 1.33 N)
S = (6.4 mm) ÷ (13.87 N)
⸫ S = 0.46 mm/N
d) We use the formula given to find the spring constant:
Spring Constant = 1/S
Spring Constant = 1 ÷ 0.46 mm/N
⸫ Spring Constant = 2.17 N/mm
e)
When there is no load on the spring, the length of the spring is 8.5 mm as seen from the graph above.
f)
g) The extension of the spring for the same load is calculated using Hooke's Law:
Force = spring constant × extension
7.0 N = 2.17 N/mm × extension
extension = 7.0 N ÷ 2.17 N/mm
⸫ extension = 3.23 mm
h) We use Hooke's Law to determine the load which would produce an extension of 4.0 mm:
Force = spring constant × extension
Force = 2.17 N/mm × 4.0 mm
⸫ Force = 8.68 N
i) The student would know that she had not taken the spring past its elastic limit when the points plotted fall along a line of best-fit. When these measurements tend to curve upward, she would have passed the elastic limit of the spring.
Example 2
The figure below shows how the length of a spring changes when loaded.
a) The extension produced by the 200 N load
b) The load that produces an extension of 1 cm
c) The load X
d) The length of the spring when a 120 N load is attached to it.
Solution
a) To find the extension produced by the 200 N load, we need to determine the Original Length of the spring. The Original Length of the spring is the length of the spring when no load is applied.
Length of Spring when 200 N force is applied to the spring = 25 cm
Length of Spring = Original Length + extension
25 cm = 15 cm + extension
extension = 25 cm - 15 cm
⸫ extension = 10 cm
b) We know a 200 N load produces an extension of 10 cm. Therefore, we can use two methods to calculate the load that produces an extension of 1 cm.
Method 1: Find the spring constant
Force = spring constant × extension
200 N = spring constant × 10 cm
spring constant = 200 N ÷ 10 cm
⸫ spring constant = 20 N/cm
For an extension of 1 cm,
Force = spring constant × extension
Force = 20 N/cm × 1 cm
⸫ Force = 20 N
Method 2: Because there is a directly proportional relation between force and extension, we can use ratios
10 cm = 200 N
Divide both sides of the equation by 10
10 cm ÷ 10 = 200 N ÷ 10
⸫ 1 cm = 20 N
Therefore, a 20 N force will produce an extension of 1 cm.
c) To find Load X, we first need to determine the extension it produced on the spring:
Length of Spring = Original Length + extension
18 cm = 15 cm + extension
extension = 18 cm - 15 cm
⸫ extension = 3 cm
We now use Hooke's Law equation to determine Load X:
Force = spring constant × extension
Load X = 20 N/cm × 3 cm
⸫ Load X = 60 N
d) First, we find the extension produced by the 120 N load when attached and then, add it to the Original Length of the Spring:
Force = spring constant × extension
120 N = 20 N/cm × extension
extension = 120 N ÷ 20 N/cm
⸫ extension = 6 cm
Therefore,
Length of Spring = Original Length + extension
Length of Spring = 15 cm + 6 cm
⸫ Length of Spring = 21 cm
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