A basketball player makes a jump shot. The 0.600-kg ball is released at a height of 2.01 m above the floor with a speed of 7.26 m/s. The ball goes through the net 3.10 m above the floor at a speed of 3.81 m/s. What is the work done on the ball by air resistance, a nonconservative force

Work done on the ball by air resistance = -5.049J

Explanation:

From law of conservation of energy

Potential energy + kinetic energy + Work of air resistance  = Potential energy + kinetic energy +

mgh + 1/2 mv^2  + W= mgh¹ + 1/2 mv¹^2

=0.6 Kg x 9.8m/s x 2.01  +1/2 x0.60 x (7.26)^2+ W= 0.6 x 9.8 x 3.10 + 1/2 x 0.6 x 3.81

=11.8188 + 15.81288 +W = 18.228 + 4.3548

=27.63168 +W =22.58283

W = 22.578-27.63168

W=-5.049 J

Work done on the ball by air resistance = -5.049J ( N egative sign indicates that  Work was in opposite direction of motion

trig 15.3 cos