Question

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?

Answer 1

The energy transferred to the spring is given by:
$U= \frac{1}{2}kx^2$
where 
k is the spring constant
x is the elongation of the spring with respect its initial length

Let's convert the data into the SI units:
$k=52.9 N/cm = 5290 N/m$
$x=40.0 cm=0.4 m$

so now we can use these data inside the equation ,to find the energy transferred to the spring:
$U= \frac{1}{2}kx^2= \frac{1}{2}(5290 N/m)(0.4m)^2=423.2 J$