When a click beetle is upside down on its back, it jumps upward by suddenly arching its back, transferring energy stored in a muscle to mechanical energy. This launching mechanism produces an audible click, giving the beetle its name. Videotape of a certain click-beetle jump shows that a beetle of mass $\mathbf{m}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}\mathbf{}}\mathbf{kg}$moved directly upward by 0.77 mm during the launch and then to a maximum heighth = 0.30 m. During the launch, what are the average magnitudes of (a) the external force on the beetles back from the floor and (b) the acceleration of the beetle in terms of g?

Mass of the beetle is,$\mathrm{m}=4.0\mathrm{x}{10}^{-5}\mathrm{kg}$

Distance

$$\begin{gathered} \mathrm{d}=0.77\mathrm{mm} \\ =7.7\times {10}^{-4}\mathrm{m} \end{gathered}$$

The maximum height h = 0.30 m

The problem deals with the Newton’s second law of motion which states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object. First, we have to find the mechanical energy $\mathit{∆}{\mathit{E}}_{\mathit{m}\mathit{e}\mathit{c}\mathit{,}\mathit{0}}$before the launch and mechanical energy $\mathit{∆}{\mathit{E}}_{\mathit{m}\mathit{e}\mathit{c}\mathit{,}\mathit{1}}$at the maximum height h . Then, by using these values, we can find the average magnitude of external force ${\mathit{F}}_{\mathbf{a}\mathbf{v}\mathbf{g}}$. Finally, by using Newton’s second law, we can find theacceleration of the beetle in terms of g.

Formula:

The change of mechanical energy is related to the external force as

localid="1661400421736" $∆E_{mec}=∆E_{mec,1}-∆E_{mec,0}$ And
localid="1661400901409" $∆E_{mec}=mgh$

$=F_{avg}d\cos \varphi$

Newton’s second law,$F_{avg}=ma$

The mechanical energy is

$∆E_{mec.0}=0$

At maximum height, the mechanical energy is given by

$∆E_{mec.1}=mgh$

The change of mechanical energy is related to the external force is given as

$$\begin{gathered} ∆E_{mec}=∆E_{mec,1}-∆E_{mec,0} \\ ∆E_{mec}=mgh \\ =F_{avg}d\cos \varphi \end{gathered}$$

Where, $F_{avg}$is average magnitude of external force on the bottle.

$$\begin{gathered} F_{avg}=\frac{mgh}{d\cos \varphi } \\ =\frac{\left(4.0\times {10}^{-6}\mathrm{kg}\right)\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\times \left(0.30\mathrm{m}\right)}{\left(7.7\times {10}^{-4}\mathrm{m}\right)\times \left(\cos 0°\right)} \\ =1.5\times {10}^{-2}\mathrm{N} \end{gathered}$$

Hence, the external force on the beetles back from the floor is,${\mathrm{F}}_{\mathrm{avg}}=1.5\times {10}^{-2}\mathrm{N}$

Dividing the above result by the mass of the bottle, we get,

$$\begin{gathered} \mathrm{a}=\frac{{\mathrm{F}}_{\mathrm{avg}}}{\mathrm{m}} \\ =\frac{\mathrm{h}}{\mathrm{d}\mathrm{cos\varphi }}\mathrm{g} \\ =\frac{\left(0.30\mathrm{m}\right)}{\left(7.7\times {10}^{-4}\mathrm{m}\right)\times \left(\cos 0°\right)}\mathrm{g} \\ =3.8\times {10}^2\mathrm{g} \end{gathered}$$

Hence, the acceleration of the beetle in terms of g is,$\mathrm{a}=3.8\times {10}^2\mathrm{g}$