Does Changing the Angle of a Mop Handle Affect the Work Done by a Janitor?

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To clean a floor, a janitor pushes on a mop handle with a force of 58.4

A. If the mop handle is at an angle of 59 above the horizontal, how much work is required to push the mop 0.55?



B. If the angle the mop handle makes with the horizontal is increased to 65 , does the work done by the janitor increase, decrease, or stay the same?

Can someone please walk me through this problem

Thanks
 
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Use

[tex]W = \int _a ^b \vec F \cdot \vec ds[/tex]

which in this case, since the force is constant, reduces to the more simple

[tex]W = \vec F \cdot \vec s[/tex]

where [tex]W[/tex] is the work, [tex]\vec F[/tex] is the force, and [tex]\vec s[/tex] is the displacement (i.e. distance). The dot between [tex]\vec F[/tex] and [tex]\vec s[/tex] is the dot product (also called the inner product) operator, since you are working with vectors.

Be careful though, don't forget you are working with vectors. The above equation can be interpreted as the work is found by multiplying only the component of the force that happens to be parallel with the displacement, with the displacement (resulting in a scalar value). Since only part of the force is parallel with the mop's movement, you'll have to fit a [tex]cos \theta[/tex] into your solution somewhere.