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:
$U= \frac{1}{2}kx^2= \frac{1}{2}(5290 N/m)(0.4m)^2=423.2 J$

Answer 2

The amount of energy required by the athlete to transferred to the spring is 423.2 J.

What is elastic potential energy?

The elastic potential energy is the energy which is stored in a stretched spring. The elastic potential energy of a spring can be found by the following formula.

$U_s=\dfrac{1}{2}kx^2$

Here, (k) is the elastic constant and (x) is the displacement from the equilibrium point.

The athlete stretches a spring an extra 40.0 cm or 0.4 m beyond its initial length, and the spring constant is 52.9 N/cm or 5290 N/m. Thus, the elastic potential energy is,

$U_s=\dfrac{1}{2}(5290)(0.4)^2\\U_s=423.2\rm \; J$

Hence, the amount of energy required by the athlete to transferred to the spring is 423.2 J.

Learn more about the elastic potential energy here;

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