Question

A basketball player makes a jump shot. The .6-kg ball is released at a height of 2 m above the floor with a speed of 7.2 m/s. the ball goes through the net 3.2 m above the floor at a speed of 4.2 m/s. what is the work done on the ball by air resistance, a non conservative force?

Answer 1

A basketball player makes a jump shot. The .6-kg ball is released at a height of 2 m above the floor with a speed of 7.2 m/s. the ball goes through the net 3.2 m above the floor at a speed of 4.2 m/s. what is the work done on the ball by air resistance, a non conservative force?

Solution

$$E _ {1} = K _ {1} + P _ {1} = \frac {m v _ {1} ^ {2}}{2} + m g h _ {1}$$ $$E _ {2} = K _ {2} + P _ {2} = \frac {m v _ {2} ^ {2}}{2} + m g h _ {2}$$

the work done on the ball by air resistance

$$A = E _ {1} - E _ {2} = \frac {m v _ {1} ^ {2}}{2} + m g h _ {1} - \frac {m v _ {2} ^ {2}}{2} - m g h _ {2}$$ $$\begin{array}{l} A = \frac {m (v _ {1} ^ {2} - v _ {2} ^ {2})}{2} + m g (h _ {1} - h _ {2}) = \frac {0 . 6 (7 . 2 ^ {2} - 4 . 2 ^ {2})}{2} + 0. 6 * 9. 8 (2 - 3. 2) \\ = 1 0, 2 6 - 7. 0 6 = 3. 2 \mathrm {J} \\ \end{array}$$