Question

A 0.50 kgbanana is thrown directly upward with an initial speed of4.00 m/sand reaches a maximum height of 0.80 m. What change does air drag cause in the mechanical energy of the banana-Earth system during the ascent?

Short answer

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

Step-by-step solution

Given data:

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, $h=h_f-h_i=0.8\mathrm{m}$

To understand the concept:

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:

$W=∆E_{mech}+∆E_{th}$

Calculate what change the air drag causes in the mechanical energy of the banana-Earth system during the ascent:

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

$W=∆E_{mech}+∆E_{th}$

In this case, W = 0 Then,

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

Since

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

Therefore,

$∆E_{mech}=0.08\mathrm{J}$

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.