KE1-KE2=W(fr)
v1= 96.6 mi/h = 43.18 m/s
v2 = 95.3 mi/h = 42.60 m/s
W(fr) =m•(v1)^2/2 - m•(v2)^2/2 =3.6 J
I am having alot of difficulty with this problem,any help would be greatly appreciated.
A baseball of mass 145 g leaves a pitcher’s hand at 96.6 mi/h (1 mi/h = 1.609km/h), but due to air resistance, it arrives at home plate 60.0 ft away (1 ft = 0.3048 m) traveling at 95.3 mi/h. Assume that the magnitude of the ball’s acceleration is constant and that the ball travels in a straight line (ignore gravity). How much work is done by friction during the flight of the ball?
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1 answer