Question

a boy holds a ball at shoulder level before throwing it. four possible paths the ball could travel and the locations of the ball at some point along those paths are shown below. for each location, select whether the potential energy stored in the ball-earth system is more, less, or the same as when the ball was held at the boy’s shoulder. click each dot on the image to select an answer.

Explanation:

To solve this, we use the formula for gravitational potential energy: \( PE = mgh \), where \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is height above a reference point (the boy’s shoulder level here).

Step 1: Analyze the top - most blue ball

The height of this ball is greater than the height of the boy’s shoulder. Since \( PE = mgh \) and \( m \) and \( g \) are constant, a greater \( h \) means greater potential energy. So, the potential energy here is more than at the shoulder.

Step 2: Analyze the blue ball on the tree (same horizontal level as shoulder)

The height of this ball is equal to the height of the boy’s shoulder (same horizontal dashed line). Since \( h \) is the same, and \( m \) and \( g \) are constant, \( PE = mgh \) will be the same. So, the potential energy here is the same as at the shoulder.

Step 3: Analyze the blue ball near the green - shirted boy (below shoulder level)

The height of this ball is less than the height of the boy’s shoulder. Since \( PE = mgh \) and \( m \) and \( g \) are constant, a smaller \( h \) means less potential energy. So, the potential energy here is less than at the shoulder.

Step 4: Analyze the blue ball on the lower path (below shoulder level)

The height of this ball is less than the height of the boy’s shoulder. Since \( PE = mgh \) and \( m \) and \( g \) are constant, a smaller \( h \) means less potential energy. So, the potential energy here is less than at the shoulder.

For the top - most ball: More
For the ball on the tree (same level as shoulder): Same
For the ball near the green - shirted boy: Less
For the ball on the lower path: Less

(Note: The exact labels of the balls can be matched visually. The key is comparing their heights to the shoulder level using \( PE = mgh \).)

Answer:

To solve this, we use the formula for gravitational potential energy: \( PE = mgh \), where \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is height above a reference point (the boy’s shoulder level here).

Step 1: Analyze the top - most blue ball

The height of this ball is greater than the height of the boy’s shoulder. Since \( PE = mgh \) and \( m \) and \( g \) are constant, a greater \( h \) means greater potential energy. So, the potential energy here is more than at the shoulder.

Step 2: Analyze the blue ball on the tree (same horizontal level as shoulder)

The height of this ball is equal to the height of the boy’s shoulder (same horizontal dashed line). Since \( h \) is the same, and \( m \) and \( g \) are constant, \( PE = mgh \) will be the same. So, the potential energy here is the same as at the shoulder.

Step 3: Analyze the blue ball near the green - shirted boy (below shoulder level)

The height of this ball is less than the height of the boy’s shoulder. Since \( PE = mgh \) and \( m \) and \( g \) are constant, a smaller \( h \) means less potential energy. So, the potential energy here is less than at the shoulder.

Step 4: Analyze the blue ball on the lower path (below shoulder level)

The height of this ball is less than the height of the boy’s shoulder. Since \( PE = mgh \) and \( m \) and \( g \) are constant, a smaller \( h \) means less potential energy. So, the potential energy here is less than at the shoulder.

For the top - most ball: More
For the ball on the tree (same level as shoulder): Same
For the ball near the green - shirted boy: Less
For the ball on the lower path: Less

(Note: The exact labels of the balls can be matched visually. The key is comparing their heights to the shoulder level using \( PE = mgh \).)