# Please see the link. Note: http://people.westminstercollege.edu/faculty/ccline/courses/phys211/hw-f07/hw-211-u10.pdf question # 10-2)

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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.**