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

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.

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  1. Answer:

    [tex]W=173.48J[/tex]

    Explanation:

    information we know:

    Total force: [tex]F=45N[/tex]

    Weight: [tex]w=100N[/tex]

    distance: [tex]4m[/tex]

    vertical component of the force: [tex]F_{y}=12N[/tex]

    ————-

    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

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