# Basic question concerning the concept of a work integral

1. Sep 18, 2012

### aftershock

Edit: I think I figured it out actually. Left must be negative by the way I introduced the x-axis meaning the force is -10 not 10 when I set up the integral.

Hey guys, this isn't a homework question but solving a hw problem led me to think about this.

Imagine I have a mass situated on the x-axis. The mass is at x=5m, I then apply a 10N force to the object until it is at x=2m.

The force is 10, the distance is 3, they're in the same direction and the angle is zero so I performed +30J of work.

But now if I evaluate it as an integral I can say the force is ∫10dx evaluated from 5 to 2 which becomes 10*2 -10*5 = -30

So the integral tells me I performed negative work How could that be?

Last edited: Sep 18, 2012
2. Sep 18, 2012

### mikeph

Your integral is missing a minus sign, since the force is -10N.