Springs and Hooke’s Law – Practice Answers #
Answer 1 [3 marks] #
Using Hooke’s Law: F = kx
Where:
- k = 25 N/m
- x = 0.08 m
F = 25 × 0.08 = 2 N
Marking points:
- Correct formula [1]
- Correct substitution [1]
- Correct answer with units [1]
Answer 2 [4 marks] #
a) Finding spring constant using F = kx
k = F/x = 2/0.05 = 40 N/m [2]
b) Finding elastic potential energy using E = ½kx²
E = ½ × 40 × (0.05)² = 0.05 J [2]
Marking points:
- Correct calculation of spring constant [2]
- Correct elastic potential energy formula and substitution [1]
- Correct final answer with units [1]
Answer 3 [4 marks] #
a) Point P represents the elastic limit of the spring [2]
This is the point where Hooke’s Law stops applying, and the spring’s behavior changes from elastic to plastic deformation.
b) After point P [2]:
- The spring enters the plastic region
- It no longer follows Hooke’s Law
- Any further stretching will cause permanent deformation
- The spring won’t return to its original length when the force is removed
Answer 4 [3 marks] #
First find k using F = kx:
4 = k × 0.1
k = 40 N/m
Then find energy using E = ½kx²:
E = ½ × 40 × (0.1)²
E = 0.2 J
- Correct calculation of spring constant [1]
- Correct use of energy formula [1]
- Correct final answer with units [1]
Answer 5 [6 marks] #
a) Using F = kx for each spring:
Spring A: 5 = k × 0.25, so k = 20 N/m
Spring B: 5 = k × 0.1, so k = 50 N/m [2]
b) Spring B is stiffer because:
- It has a larger spring constant (50 N/m vs 20 N/m)
- It extends less for the same force [2]
c) For 0.15m extension:
Spring A: F = 20 × 0.15 = 3 N
Spring B: F = 50 × 0.15 = 7.5 N
Spring B requires more force (7.5 N) [2]
