Work done by the automatic pitcher as the angle changes

[bostampa2007]

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for this problem, we're supposed to use energy=1/2 mv2 to show that work done does not change. I don't understand why we can't use Work=Fdcos@ which would different answer for different angle.

bottom line: to calculate work done: how do we choose 1/2mv2 over the other in any given problem (given both could be used)?

 

[free_radical]

i cant answer your particular quesiton, but ill tell you EK's general strategy, which i found useful.

theres three possible ways you can answer the question - "how much work was done"?

- Fdcos(theta)

- the difference in gravitational potential energy

- everything else

the second you see something of a given mass change heights, you want to plug in the difference in grav. potential energy. if you see a force applied to an object acting over a distance, you can plug in Fdcos(theta). calculating the difference in kinetic energy......i dont know exactly when youd use that.

what exact example were you referring to? maybe it would help to analyze the question

 

[Hindiana_Jones]

You don't use W=Fdcos because changing the angle of the pitcher to the ground will not change the angle between the force (the pressure in the cannon) and the d(the length of the cannon tube). I think F and D are parallel to each other.

 

[mrmilad]

This problem may be related to centripetal motion and uniform circular motion. As the automatic pitcher swings the ball through Q, the centripetal force
Fc=mass * v^2 / r , the distance from the center for this make shift circle is constant even though theta may be changing, thus the work done would be 0; just as the work done on the moon by the earth is also equal to 0. I am not sure if this would apply completely because you didn't give the question stem, but if it is not uniform circular motion, then you may want to use gravitational potential energy. Using the theta to find the change in the Y direction as the angle changes, then use W=mg delta h.