Calculating work with variable force

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SUMMARY

The discussion focuses on calculating work done by variable forces, specifically in the context of a physics problem from MIT's course materials. The key equation referenced is W = ∫(F_y dy), where the work done is attributed solely to gravitational force (mgh) despite the presence of tension in the string. The conclusion is that tension does not perform work on the rock since it is always perpendicular to the motion, confirming that only gravity contributes to the work done in this scenario.

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forester404
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Homework Statement


Hi.
I am trying to learn some physics with MIT free resources (just out of personal interest),
and this is where i got stuck:
on section 14-13, in
http://ocw.mit.edu/courses/physics/8-01-physics-i-fall-2003/lecture-notes/binder13.pdf"
I am a little puzzled on how the work is calculated in the given solution - mgh,
where h is delta y between the 2 points. if i got it right, this calculation assumes
that the net force on y direction is mg and is constant throughout the path, but it seems
to me like the tension in the string is larger than 0 in every point along the path
except the very top of the circle (where it's 0 by porblem definition), and
always has some y component, so i don't understand how the net force in y direction
is a constant mg. what am i missing here ?

Homework Equations



W = integral from a to b (Fydy)

(sorry but i couldn't handle the LaTex renderer here...)

The Attempt at a Solution

 
Last edited by a moderator:
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The tension in the string is always perpendicular to the motion, thus it does no work on the rock. Only gravity does work on the rock.
 
Thanks for the quick reply !
 

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