Calculating Work Done by a Batter on a Baseball

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To calculate the work done by a batter on a baseball, the change in kinetic energy (W = ΔKE) is used, where the final kinetic energy (KE_f) is 200 joules and the initial kinetic energy (KE_i) is 12.5 joules. The correct approach is to subtract the initial kinetic energy from the final kinetic energy, resulting in 187.5 joules of work done. There is confusion regarding whether to add the kinetic energies, but work is a scalar quantity and should be calculated as the difference. The discussion clarifies that the teacher's perspective may conflate work with impulse, affirming the validity of the subtraction method. Understanding these principles is crucial for accurately calculating work in physics.
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Homework Statement


A baseball with a mass of 1kg is thrown with a velocity of -5 m/s. A batter hits the baseball and sends the ball into the outfield with a velocity of 20 m/s. What is the amount of work done by the batter?


Homework Equations


<br /> W = \Delta KE = KE_{f} - KE_{i} <br />
<br /> KE = \frac{1}{2} m v^2<br />

The Attempt at a Solution


After substituting the values into the formula for kinetic energy, I determined that the final kinetic energy of the baseball is 200 joules and the initial kinetic of the ball is 12.5 joules. Would I then simply subtract 12.5 joules from 200 joules to arrive at my final answer?
My teacher insists that the two values need to be added in order to arrive at the work done, and he tells me that even though the work done to change the ball's velocity from -5 m/s to 0 m/s is negative, the magnitudes of the values need to be added to find the total work. Could anybody shed some light on this?
 
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It's hard to argue successfully with a teacher.

However, you are correct! Your analysis spot-on (absolutely right).

Work is a scalar property.

Sam Snyder, PhD in physics, if that helps convince him.
 
I think your teacher is confusing work with impulse.
 
Your method is totally fine.
 
Thank you all for your help!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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