Question 3
ElasticityA string of length 5m is extended by 0.04m when a load of 0.8kg is suspended at the end. How far will it extend if a force of 16N is applied? [g = 10ms−2]
- A.0.04m
- B.0.12
- C.0.01m
- D.0.08m
Show the answer and explanation
Answer: D (0.08m)
Explanation
The load weighs 0.8 × 10 = 8 N. By Hooke's law extension is proportional to force, so 16 N (double) gives 2 × 0.04 = 0.08 m.