Question

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.

Answers

  1. Answer:

    W=173.48J

    Explanation:

    information we know:

    Total force: F=45N

    Weight: w=100N

    distance: 4m

    vertical component of the force: F_{y}=12N

    ————-

    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: F_{x}=Fcos\theta

    vertical component: F_{y}=Fsen\theta

    but from the given information we know that F_{y}=12N

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

    Fsen\theta =12N

    and we know the force F=45N, thus:

    45sen\theta=12

    now we clear for \theta

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

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

    F_{x}=Fcos\theta

    F_{x}=45cos(15.466)\\F_{x}=43.37N

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

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

    the work done is W=173.48J

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