a basketball player makes a jumpshot. the .600-kg ball is released at a height o

Question

a basketball player makes a jumpshot. the .600-kg ball is released at a height of 2.00 m above the floor with a speed of 7.20 m/s. the ball goes through the net 3.10 m above the floor at a speed of 4.20 m/s. what is the work done on the ball by air resistance, a nonconservative force? a basketball player makes a jumpshot. the .600-kg ball is released at a height of 2.00 m above the floor with a speed of 7.20 m/s. the ball goes through the net 3.10 m above the floor at a speed of 4.20 m/s. what is the work done on the ball by air resistance, a nonconservative force?

Explanation / Answer

mass of the ball m = 0.6 kg

initial height h = 2.0 m

final height h ' = 3.1 m

initial speed v = 7.2 m / s

final speed v ' = 4.2 m / s

Let the work done on the ball by air resistance W = ?

we know from law of conservation of energy ,

total energy at height h + work done by air = total energy at height h '

   mgh + ( 1/ 2) mv^ 2 + W = mgh ' + ( 1/ 2) mv'^ 2

11.76 + 15.552 + W = 18.228 + 5.292

W = -3.792 J

here negative sign indicates it work opposite direction to motion

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