Answer:
[tex]W_{drag} = 4.223\,J[/tex]
Explanation:
The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:
[tex]U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}[/tex]
The work done on the ball due to drag is:
[tex]W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})[/tex]
[tex]W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})[/tex]
[tex]W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}][/tex]
[tex]W_{drag} = 4.223\,J[/tex]
Answer:
W = -4.22 J
Explanation:
Given
m = 0.599 kg
vi = 7.05 m/s
yi = 2.18 m
vf = 4.19 m/s
yf = 3.10 m
We apply the equations of the Principle of Energy Conservation and the Work-Energy Theorem
W = Ef - Ei
W = (Kf + Uf) - (Ki + Ui)
W = (m/2)(vf² - vi²) + mg(yf - yi)
W = (0.599 kg/2)((4.19 m/s)² - (7.05 m/s)²) + (0.599 kg)(9.81m/s²)(3.10 m - 2.18 m)
W = -4.22 J
