91影视

Mass of the banana, m = 0.50 kg

Initial speed of the banana,$v_i=4\mathrm{m}/\mathrm{s}$

Maximum height reached by the banana, $h=h_f-h_i=0.8\mathrm{m}$

Using the formula for workdone on a system by external force, find thechange caused by the air drag in the mechanical energy of the banana-Earth system during the ascent.

Formula:

$W=鈭�E_{mech}+鈭�E_{th}$

Work done on a banana-Earth system by external force is given by,

$W=鈭�E_{mech}+鈭�E_{th}$

In this case, W = 0 Then,

$$\begin{gathered} 0=鈭�E_{mech}+鈭�E_{th} \\ 0=鈭�KE+鈭�PE+鈭�E_{th} \\ 0=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2+mg(h_f-h_i)+鈭�E_{th} \\ 0=0-\frac{1}{2}(0.50){\left(4\right)}^2+(0.50)(9.8)(0.8)+鈭�E_{th} \\ 鈭�E_{th}=0.08\mathrm{J} \end{gathered}$$

Since

$$\begin{gathered} 鈭�E_{mech}+鈭�E_{th}=0 \\ 鈭�E_{mech}=-鈭�E_{th} \\ 鈭�E_{mech}=\left|鈭�E_{th}\right| \end{gathered}$$

Therefore,

$鈭�E_{mech}=0.08\mathrm{J}$

Hence, the change caused by the air drag in the mechanical energy of the banana-Earth system during the ascent is equal to 0.08 J.