QUESTION IMAGE
Question
a group of skydivers gains speed as they fall together.
a graph relating the groups kinetic energy in kilojoules (kj) and their speed in meters per second ($\frac{m}{s}$) is shown below.
use the pattern on the graph to complete the prediction.
- when the groups speed is 40 $\frac{m}{s}$, their kinetic energy will be \boxed{} kj.
Step1: Identify the relationship type
The graph of kinetic energy (KE) vs. speed (v) is a curve that looks like a quadratic (or power - law) relationship. Recall that the formula for kinetic energy is $KE=\frac{1}{2}mv^{2}$, which is a quadratic relationship (KE is proportional to $v^{2}$). Let's check the given points. When $v = 20\frac{m}{s}$, $KE=80\space kJ$. Let's verify the proportionality. If $KE\propto v^{2}$, then $\frac{KE_1}{v_1^{2}}=\frac{KE_2}{v_2^{2}}$.
Step2: Set up the proportion
Let $v_1 = 20\frac{m}{s}$, $KE_1 = 80\space kJ$, $v_2=40\frac{m}{s}$, and $KE_2$ be the unknown kinetic energy. Since $KE\propto v^{2}$, we have $\frac{KE_1}{v_1^{2}}=\frac{KE_2}{v_2^{2}}$. Substitute the known values: $\frac{80}{20^{2}}=\frac{KE_2}{40^{2}}$.
Step3: Solve for $KE_2$
First, calculate $20^{2}=400$ and $40^{2} = 1600$. The equation becomes $\frac{80}{400}=\frac{KE_2}{1600}$. Cross - multiply: $400\times KE_2=80\times1600$. Then, $KE_2=\frac{80\times1600}{400}$. Simplify: $80\times\frac{1600}{400}=80\times4 = 320$.
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