Internal Work of a Ball Shot Through a Sand Bag

AI Thread Summary
The problem involves a ball shot straight down with a velocity of 44 m/s and a mass of 0.09 kg, encountering a sand bag after falling 18 meters and penetrating it by 0.5 meters. The internal work is calculated using the formula w = -mgh - 0.5mv^2. The vertical distance for gravity's work should include both the fall and the penetration, totaling 18.5 meters. Clarification was provided that gravity continues to act throughout the entire distance, including the 0.5 meters inside the bag. The discussion emphasizes understanding the total distance over which gravity performs work on the ball.
gamarrapaul
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in the problem they give you V=44m/s, mass=.09 kg , and a vertical distance=18m
the problem states that a ball is shot straight down and at the distance 18m it encounters a sand bag the ball goes through the sand bag .5 m. they ask what is the internal work neglecting friction?
for this problem I have

-w=-mgh-.5mv^2

now I am confused for h do I use 18 m or 18.m ?
 
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You use the distance over which gravity was doing work on the ball. Which is that?
 
the distance that gravity is doing work on the ball will be 18m ??
 
The distance gravity is doing work on the ball is the vertical distance that the ball moves. Gravity doesn't shut off inside the bag.
 
oh so the distance will be 18.5m I thought that .5 was the distance that the internal force is acting on the object Work = F*d
 
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