WAEC Physics 2017 Paper

Concept:Elastic potential energy stored in a spring is equal to the work done in stretching it, given by $\frac{1}{2} k x^2$.Explanation:The spring has stiffness constant $k = 2000\ \mathrm{N\,m^{-1}}$.The extension is $4\ \mathrm{cm}$, which must be converted to metres:$x = 4\ \mathrm{cm} = 0.04\ \mathrm{m}$.Energy stored is calculated using the formula:$$E = \frac{1}{2} k x^2$$Substitute the given values:$$E = \frac{1}{2} \times 2000 \times (0.04)^2$$First, $(0.04)^2 = 0.0016$.Then, $2000 \times 0.0016 = 3.2$.Therefore, $E = \frac{3.2}{2} = 1.60\ \mathrm{J}$.Answer:The energy stored in the spring is $1.60\ \mathrm{J}$.Correct option: B. $1.60\ \mathrm{J}$.