locust

[logistic_guy]

\(\displaystyle \bold{(a)}\) A \(\displaystyle 3.0\)-\(\displaystyle \text{g}\) locust reaches a speed of \(\displaystyle 3.0 \ \text{m/s}\) during its jump. What is its kinetic energy at this speed? \(\displaystyle \bold{(b)}\) If the locust transforms energy with \(\displaystyle 40\) percent efficiency, how much energy is required for the jump?

 

[khansaheb]

\(\displaystyle \bold{(a)}\) A \(\displaystyle 3.0\)-\(\displaystyle \text{g}\) locust reaches a speed of \(\displaystyle 3.0 \ \text{m/s}\) during its jump. What is its kinetic energy at this speed? \(\displaystyle \bold{(b)}\) If the locust transforms energy with \(\displaystyle 40\) percent efficiency, how much energy is required for the jump?

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[logistic_guy]

\(\displaystyle \bold{(a)}\)

The kinetic energy equation is:

\(\displaystyle K = \frac{1}{2}mv^2\)

Then,

\(\displaystyle K = \frac{1}{2}(0.003)3^2 = \textcolor{blue}{13.5 \ \text{mJ}}\)

 

[logistic_guy]

\(\displaystyle \bold{(b)}\) If the locust transforms energy with \(\displaystyle 40\) percent efficiency, how much energy is required for the jump?

In this case this little locust will need to do more work to make the same jump.

Let \(\displaystyle E\) be his new energy, then

\(\displaystyle E = \frac{K}{0.4} = \frac{0.0135}{0.4} = 0.03375 \ \text{J} = \textcolor{blue}{33.75 \ \text{mJ}}\)