A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a distance of 4 m. Given that the vertical component of the pulling force is 12 N, calculate the work done by the force in moving the crate.

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: [tex]F_{x}=Fcos\theta[/tex]

vertical component: [tex]F_{y}=Fsen\theta[/tex]

but from the given information we know that [tex]F_{y}=12N[/tex]

so, equation these two [tex]F_{y}=Fsen\theta[/tex] and [tex]F_{y}=12N[/tex]

[tex]Fsen\theta =12N[/tex]

and we know the force [tex]F=45N[/tex], thus:

[tex]45sen\theta=12[/tex]

now we clear for [tex]\theta[/tex]

[tex]sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466[/tex]

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

[tex]F_{x}=Fcos\theta[/tex]

[tex]F_{x}=45cos(15.466)\\F_{x}=43.37N[/tex]

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

[tex]W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J[/tex]

the work done is W=173.48J