Calculating work with variable force

In summary, The conversation discusses a question about calculating work in a physics problem and the confusion surrounding the net force in the y direction. The solution assumes that the tension in the string is 0 except at the top of the circle, where it is defined as 0 in the problem. The tension is always perpendicular to the motion and does not do any work on the rock, leaving only gravity to do work.
  • #1
forester404
2
0

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

 
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  • #2
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.
 
  • #3
Thanks for the quick reply !
 

1. How do you calculate work with variable force?

To calculate work with variable force, you need to use the formula W = ∫F(x)dx, where F(x) represents the variable force and dx is the change in position. This formula is based on the concept of integration in calculus.

2. What is the unit of measurement for work with variable force?

The unit of measurement for work with variable force is joules (J). This is the same unit used to measure other forms of energy.

3. Can you give an example of calculating work with variable force?

Sure, let's say you are pushing a box with a variable force that changes from 10N to 20N as the box moves 5 meters. The work done would be W = ∫F(x)dx = ∫(10x+5)dx = 125J.

4. How does the angle of the force affect the calculation of work with variable force?

The angle of the force does not affect the calculation of work with variable force. The formula W = ∫F(x)dx takes into account the direction of the force, so the angle does not need to be considered separately.

5. Can the work done with variable force ever be negative?

Yes, the work done with variable force can be negative. This happens when the force and displacement are in opposite directions. In this case, the work done is considered to be negative because the force is acting against the direction of motion.

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