Work Done by Force Field (Part 2)

A particle in the x y xy -plane is acted upon by a force field described by the equation below: F = x ı ^ + y ȷ ^ . \large{\vec{F} = x \hat{\imath} + y \hat{\jmath}}. How much work does the force field do on the particle if the particle makes one complete revolution over a unit circle centered on the origin?


Notes:

  • ı ^ \hat{\imath} and ȷ ^ \hat{\jmath} denote unit vectors in the x x and y y directions, respectively.
  • Give your answer to 3 decimal places.


The answer is 0.000.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Aareyan Manzoor
Jun 5, 2017

notice that the force is conservative since F x y = F y x \dfrac{\partial F_x}{\partial y}=\dfrac{\partial F_y}{\partial x} it follows that the work done on a closed path is zero.

Tom Engelsman
Apr 5, 2017

A line integral approach!

Nice! I'm thinking of posting another which is a slight variation on this one.

Steven Chase - 4 years, 2 months ago

Log in to reply

Thanks, Steven :)

tom engelsman - 4 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...