Answer
The amount of energy dissipated by air drag is $5.4~J$
Work Step by Step
We can find the initial energy:
$E_1 = K_1+U_1$
$E_1 = \frac{1}{2}mv_1^2+mgh_1$
$E_1 = \frac{1}{2}(0.60~kg)(7.6~m/s)^2+(0.60~kg)(9.80~m/s^2)(2.0~m)$
$E_1 = 29.1~J$
We can find the final energy:
$E_2 = K_2+U_2$
$E_2 = \frac{1}{2}mv_2^2+mgh_2$
$E_2 = \frac{1}{2}(0.60~kg)(4.5~m/s)^2+(0.60~kg)(9.80~m/s^2)(3.0~m)$
$E_2 = 23.7~J$
The amount of energy dissipated by air drag is the difference between the initial energy and the final energy:
$E_1- E_2 = 29.1~J - 23.7~J = 5.4~J$
The amount of energy dissipated by air drag is $5.4~J$
