Chapter 6, Problem 77GP

To determine

a. The kinetic energy of the locust at the given speed.

b. The energy required for jump at the given rate.

Answer to Problem 77GP

Solution:

a. The kinetic energy of the locust is $1.35\times {10}^{-2}\text{J}$ .

b.The energy required for jump is $0.039\text{J}$ .

Explanation of Solution

Given:

The mass of the locust is:

  $\begin{array}{l}m=3.0\text{g}\\ m=3.0\times {10}^{-3}\text{kg}\end{array}$

The speed of the locust is $v=3.0\frac{\text{m}}{\text{s}}$

Formula used: The kinetic energy expression is $\frac{1}{2}m{v}^2$ .

Where, m is the mass and vis the speed.

Calculation:

a

The kinetic energy of the locust at this given speed is:

  $\begin{array}{l}KE=\frac{1}{2}m{v}^2\\ $ KE=\frac{1}{2}\times 3.0\times {10}^{-3}\times {(3.0)}^2\text{J}$KE=13.5\times {10}^{-3}\text{J}\\ KE=1.35\times {10}^{-2}\text{J}\end{array}$

b

The energy transformation efficiency of the locust is 35%. Therefore, 35% of the locust’s total energy is converted to kinetic energy.

Therefore,

  $\begin{array}{l}35\%\text{of}E=1.35\times {10}^{-2}\text{J}\\ E=\frac{100}{35}\times 1.35\times {10}^{-2}\text{J}\\ E=3.9\times {10}^{-2}\text{J}\\ E=0.039\text{J}\end{array}$

Conclusion:

a. The kinetic energy of the locust is $1.35\times {10}^{-2}\text{J}$ .

b.The energy required for jump is $0.039\text{J}$ .