An athlete stretches a spring an extra 40.0 cm beyond its initial length.How much energy has he transferred to the spring,if the spring constant is 52.9 N/cm?
A) 423 J
B) 4230 kJ
C) 423 kJ
D) 4230 J

Question

Quizplus · Accepted Answer

The energy transferred to the spring can be calculated using the formula for the potential energy stored in a spring: $E = \frac{1}{2}kx^2$, where $E$ is the energy in joules, $k$ is the spring constant in N/m, and $x$ is the displacement in meters. Given $k = 52.9 \, \text{N/cm} = 5290 \, \text{N/m}$ (since $1 \, \text{N/cm} = 100 \, \text{N/m}$) and $x = 40.0 \, \text{cm} = 0.4 \, \text{m}$, the energy transferred is $E = \frac{1}{2} \times 5290 \times (0.4)^2 = 423 \, \text{J}$.

An Athlete Stretches a Spring an Extra 40