A 75 gFrisbee is thrown from a point 1.1 mabove the ground with a speed of 12 m/s.When it has reached a height of 2.1 m, its speed is 10.5 m/s. What was the reduction in E_mec of the Frisbee-Earth system because of air drag?

The mass of a Frisbee is$m=0.075\mathrm{kg}$

The initial speed of Frisbee,$v_1=12\mathrm{m}/\mathrm{s}$

The final speed of Frisbee,$v_2=10.5\mathrm{m}/\mathrm{s}$

The height of Frisbee,$h_1=1.1\mathrm{m}$

The final height of Frisbee,role="math" localid="1661232990159" $h_2=2.1\mathrm{m}$

Use the equation of change in mechanical energy to solve this problem. Mechanical energy is the sum of kinetic and potential energy

Formulae are as follow:

$$\begin{gathered} ∆E_{mec}=\left(K_1+U_1\right)-\left(K_2+U_2\right) \\ K=\frac{1}{2}{\mathrm{mv}}^2 \\ U=mgh \end{gathered}$$

where, K is kinetic energy, Uis potential energy, m is mass, v is velocity, g is an acceleration due to gravity, h is height and$∆E_{mec}$ is mechanical energy.

The equation for change in mechanical energy is,

$∆E_{mec}=\left(K_1+U_1\right)-\left(K_2+U_2\right)$

Using $K=\frac{1}{2}{\mathrm{mv}}^2$and $U=mgh$,

$$\begin{gathered} ∆E_{mec}=\frac{1}{2}m\left(v_1^2-v_2^2\right)+mg\left(h_1-h_2\right) \\ ∆E_{mec}=\frac{1}{2}\times 0.075\times \left({12}^2-10.5^2\right)+0.075\times 9.81\times \left(1.1-2.1\right) \\ ∆E_{mec}=1.27-0.74 \\ ∆E_{mec}=0.53\mathrm{J} \end{gathered}$$

Therefore, the amount of decrease in mechanical energy of the Frisbee-Earth system is $∆E_{mec}=0.53\mathrm{J}$.