Please see the link. Note: question # 10-2)

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to remember that the work done by the applied force is the dot product of the vector force and displacement vector,such that:

`W = bar F*bar d`

Using the definition of dot product, yields:

`W = |barF|*|bar d|*cos(hat(bar F, bar d))`

magnitude of force F = ` |bar F|`

magnitude of displacement d = `|bar d|`

angle between force and displacement = `(hat(bar F, bar d)) = theta`

Considering the case a), you may notice that the direction of force is perpendicular to the direction of movement, hence, theta = 90^o, thus, `cos theta = cos 90^o = 0` and the work done is  `W = |barF|*|bar d|*0 = 0` .

Considering the case b), the angle between the direction of force and the movement direction is found in quadrant 1, where the value of cosine function is positive, thus `cos theta>0` , such that:

`W = |barF|*|bar d|*cos theta > 0`

Considering the case c),you may notice that the direction of force is opposite to the direction of movement,hence `theta = 180^o` and `cos theta = cos 180^o = -1` , such that:

`W = |barF|*|bar d|*(-1) = - |barF|*|bar d < 0`

Hence, evaluating the case with the most positive work done yields that case b) satisfies this condition, while the case c) is the case with the most negative work done.

We’ve answered 319,639 questions. We can answer yours, too.

Ask a question