JAMB Physics 2024 Paper

Concept:Hooke's law states that the extension of a spring is directly proportional to the applied force, as long as the elastic limit is not exceeded.Explanation:For the same spring, the spring constant $k$ remains unchanged.From Hooke's law, $F = ke$, so $\frac{F_1}{e_1} = \frac{F_2}{e_2}$.Given: $F_1 = 10\,\text{N}$, $e_1 = 0.02\,\text{m}$, and $F_2 = 40\,\text{N}$.Substitute the values: $\frac{10}{0.02} = \frac{40}{e_2}$.Solve for $e_2$: $e_2 = \frac{40 \times 0.02}{10} = 0.08\,\text{m}$.This value is the extension produced by the $40\,\text{N}$ force, not the total length.The natural length of the spring is $1\,\text{m}$.Therefore, new length $= 1 + 0.08 = 1.08\,\text{m}$.Answer:The correct option is C. $1.08\,\text{m}$.