Question 15
ElasticityIf a load of 1 kg stretches a cord by 1.2cm, what is the force constant of the cord? [g ≈ 10ms−2]
- A.866Nm−1
- B.833Nm−1
- C.769Nm−1
- D.667Nm−1.
Show the answer and explanation
Answer: B (833Nm−1)
Explanation
k = F/e = (1 × 10) ÷ 0.012 ≈ 833 N m−1.