You drop a 5 kg bowling ball from 2 meters above a trampoline.When it bounces, it only bounces up 1 meter. How much of the ball's energy was turned into thermal energy when it hit the trampoline. 50 Joules 1 Joule 25 Joules 10 Joules

## Step 1<br />The problem involves the principle of conservation of energy, which states that energy cannot be created or destroyed, only transformed from one form to another. In this case, the bowling ball is dropped from a height of 2 meters, which means it has potential energy. When it hits the trampoline, some of this potential energy is converted into kinetic energy as the ball bounces back up, but not all of it. The ball only bounces back up 1 meter, which means it has lost some of its energy.<br /><br />## Step 2<br />The potential energy of the ball when it was dropped can be calculated using the formula:<br />### \( PE = mgh \)<br />where \( m \) is the mass of the ball, \( g \) is the acceleration due to gravity, and \( h \) is the height. Substituting the given values, we get:<br />### \( PE = 5kg * 9.8m/s^2 * 2m = 98 Joules \)<br /><br />## Step 3<br />The potential energy of the ball when it bounces back up 1 meter can be calculated using the same formula:<br />### \( PE = mgh \)<br />Substituting the given values, we get:<br />### \( PE = 5kg * 9.8m/s^2 * 1m = 49 Joules \)<br /><br />## Step 4<br />The difference between the potential energy when the ball was dropped and when it bounced back up is the amount of energy that was turned into thermal energy. This can be calculated as:<br />### \( 98 Joules - 49 Joules = 49 Joules \)