Question

The batter swings and misses the 40 m/s (90 mph) fastball, and the ball (mass 150 grams) ends up at rest in the catcher's mitt. How much work does the catcher perform on the ball

Answer 1

The work done by the catcher is 120 J.

Step-by-step explanation:

Given;

velocity of the fastball, v = 40 m/s

mass of the fastball, m = 150 g = 0.15 kg

Based on work-energy theorem, the work done by the catcher is equal to the kinetic energy of the fastball.

The kinetic energy of the fastball is given as;

K.E = ¹/₂mv²

K.E = ¹/₂ x 0.15 x 40²

K.E = 120 J

Therefore, the work done by the catcher is 120 J.