Work on a baseball being caught by a catcher

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The discussion revolves around a physics problem involving a baseball caught by a catcher. The key points include calculating the kinetic energy of the ball, determining the work done by the catcher's glove, and finding the average stopping force. Participants express confusion about applying the work-energy theorem and the implications of negative work. The conversation emphasizes understanding the direction of forces and displacement, clarifying that the stopping force acts opposite to the ball's motion, which is crucial for solving the problem accurately. Mastery of the signs in physics is highlighted as essential for understanding these concepts.
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Homework Statement


A baseball of mass 250g (0.25kg), pitched with a speed of 40m/s, is caught by a catcher whose glove moves backward 0.25m while stopping the ball.

a) What is the kinetic energy of the ball? (solved)
b) How much work did the catcher's glove do on the ball?
c) What was the average stopping force on the ball?

Given: mass=0.25kg; speed=40m/s; distance=0.25m

Homework Equations


Ek = 1/2(m)(v^2)
Work = F x d

The Attempt at a Solution


I am completely stumped, I'm trying for the life of me to figure out how to find the Applied Force on the ball with what I've been given this unit. I'm not sure how to solve for the Force Applied without two of the other variables given for the work equation.

As for question C, what are they asking for? I'm unsure of what they mean by "stopping force".
 
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If you are not familiar with the work-energy theorem, use the kinematic equations to solve for the deceleration of the ball, then apply Newton 2.
 
Is the theorem Work = Ekfinal - Ekinitial? If the initial kinetic energy = 5J, the final should be 0Jm because you're multiplying by 0 m/s. This gives me -5J which throws me off because I don't know if I should be taking the absolute value or not.
 
Symon said:
Is the theorem Work = Ekfinal - Ekinitial? If the initial kinetic energy = 5J, the final should be 0Jm because you're multiplying by 0 m/s. This gives me -5J which throws me off because I don't know if I should be taking the absolute value or not.
Work can be either positive or negative, depending on the direction of the force and the displacement.
 
Ah I understand, so when they ask for me to find "the average stopping force" are they asking me to re-arrange the W=F x d formula to solve for the force?

I'm really sorry about all the stupid questions I'm just trying to wrap my head around everything.
 
Symon said:
Ah I understand, so when they ask for me to find "the average stopping force" are they asking me to re-arrange the W=F x d formula to solve for the force?
Yes, but remember it's W=F x d x cos theta, where theta is the angle betwen the force and displacemnt vectors. What does the minus sign imply?
 
Does it imply a change in the displacement vector?
 
Symon said:
Does it imply a change in the displacement vector?
Imagine that I am the pitcher and you are the catcher. When you catch my pitched baseball, your glove moves (displaces) backward. Which way does the stopping force of your glove on the baseball act: forwards, or backwards? One of the keys to mastering Physics is mastering the plus and minus sign.
 
I would say the stopping force acts in the opposite direction that the ball is thrown, whilst the glove moves forewards in the same direction as the ball. Correct? Or have i gotten myself all backwards haha.
 
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Symon said:
I would say the stopping force acts in the opposite direction that the ball is thrown, whilst the glove moves forewards in the same direction as the ball. Correct? Or have i gotten myself all backwards haha.
You've got it right.:biggrin:
That's why the work is negative.
 
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