Question

Under what conditions are the work done i) Maximum and ii) Minimum?

Answer 1

Hint: The force must be applied for the work-done and it should not be zero
When the angle is \[90^\circ \] the work done will be minimum here, force and displacement are perpendicular to each other.
Work done will be maximum if the angle is \[0\] or \[180\] degrees, i.e force, and displacement is parallel to each other.

Complete step by step answer:
Work done \[W = FS\cos \theta \]
\[W\] is positive when \[{\text{ }}0^\circ < \theta < 90^\circ \]
\[W = 0\;\] [when \[\theta = 90^\circ \] ]
\[W\] is negative when\[{\text{ }}0^\circ < \theta < 90^\circ \].
Work done is maximum when the angle between force and displacement is zero,
\[W = FS\cos 0\]
Here, \[cos0 = 1\]
And,
\[W = FS\]
It means maximum work.
work done is minimum when the angle between force and displacement is \[90^\circ \]
\[W = FS\cos 90\]
Here, \[cos90 = 0\]
now,
\[W = 0\]
and here,
\[W = 0\]
minimum work done at \[90^\circ \]
Finally, the minimum work done will be \[0\] (i.e; when \[\theta = 90^\circ \] ).
The maximum work done at $0^\circ $.

Note:
Work is done when the object travels a certain distance in a particular direction after the force applied to it. Work done will be maximum when \[cos\theta \] is maximum, at \[\theta = 0^\circ \] and \[cos\theta \] =\[{\text{1}}\] . \[W\] refers to the work done in the opposite direction of the displacement which is negative. here, it doesn't mean the minimum work-done. The negative sign implies that it is in the opposite direction. This is a common misconception here. When we reverse the sign convention of the directions of motion and force and then observe the work done, you'll find that the positive work done, which you got in the first case, will be negative in the reversed case. It's a matter of the convention you take.