<p>This problem is #5 on the course description for mechanics. the correct answer is 16j but how do i get that?</p>
<p>The constant force F with components Fx = 3 N and Fy = 4 N, shown above, acts
on a body while that body moves from the point P (x = 2 m, y = 6 m) to the point
Q (x = 14 m, y = 1 m). How much work does the force do on the body during this
process?</p>
<p>never mind i solved it . you have to keep the x and y components separate and calculate the work done in each direction, then add them</p>
<p>Just another thing to keep in mind:</p>
<p>Work = Force * cos(t) * change in distance</p>
<p>W = Fcos(t)R</p>
<p>where t is the angle between F and R</p>
<p>So F is a magnitude 5 vector with an angle of about 53.1301 degress from the x axis</p>
<p>The slope of the line of displacement is (6-1)/(2-14) = -5/12
The inverse tangent of the slope is -22.6199 degrees</p>
<p>So T (the angle of difference between the force and the displacement is difference of the two values (53.1301 - (-22.6199)) = 75.75 degrees</p>
<p>The distance of displacement can be found using the pythagorean theorem (a.k.a. distance formula) and is 13 units long exactly</p>
<p>so cos(t) * distance = cos(75.75) * 13 = 3.1999999 units</p>
<p>Now idk if you want work as a scalar or vector quantity but your given the force has magnitude 5, Fx = 3, Fy = 4, assuming we want a vector quantity</p>
<p>Then the total work done is about 16 or so</p>