Calculating Work Done by a Batter on a Baseball

  • Thread starter Thread starter darkley
  • Start date Start date
  • Tags Tags
    Energy
AI Thread Summary
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.
darkley
Messages
2
Reaction score
0

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?
 
Physics news on Phys.org
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!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top