If work is zero, when is force nonzero?

2 Answers
Feb 18, 2017

Normally, when we talk about force and displacement, we refer to the definition of work in a one-dimensional representation:

#w = -vecFDeltavecx#

where work is negative with respect to the worker who is say, pushing a box.

One can specify that this force is parallel to the displacement, i.e. #vecF = vecF_"||" = vecFcostheta#. When #theta = 0^@#, the work is pointing exactly in the direction of the displacement.

On the other hand, if we imagine a force entirely perpendicular to the desired displacement, there is no component of the force that causes the displacement.

#vecF_(_|_) = vecFsin(0^@) = vecFcos(90^@)#

(since #sin(x + pi/2) = cos(x)#.)

When #theta = 90^@#, one pushes perpendicular to the desired displacement, but #sin0^@ = cos90^@ = 0#.

So, I don't think such a force is nonzero.

Feb 18, 2017

Force and displacement are perpendicular each other when work done is zero.


If displacement and force are perpendicular to each other, work done is zero.
#=>theta=90^o# so # W=0#
Work done is zero i.e., force and displacement are perpendicular to each other is verified in the following cases:
a) When a body is in circular motion no work is done by centripetal force.
b) In pulling a body on a horizontal surface, no work is done against gravitational force.
c) If a man with a load on his head walks horizontally no work is done against gravitational force.