<p>Is work calculated by just multiplying the applied force and the distance? Say there's a box on the floor. A person applies 10N to move the box 1 meter. But the friction force is 5 N. So is the total work 10J or 5J? do you subtract the friction force from the applied force?</p>
<p>The term work is only applicable to a certain “source”. In your case there can be a work done by the force of 10N and that done by the friction force. In both cases the work is Force*distance (cosine of the angle needs to be added if force is not parallel to the plane of movement). If force is applied in the direction of the body’s movement, work is positive. Otherwise, its negative. The total work is basically the net work of all forces acting on a body.</p>
<p>Total Work= the sum of the work done by each force (on the object)
Wfa=Fdcos(360)=Fd
My teacher was very anal about the derivations, aghhhhh</p>
<p>Work = F.d (dot product of vectors) = FdcosQ Where Q is the angle between F and d</p>
<p>Work done by man = 10<em>1 = 10J
Work done by friction = 5</em>(-1) = - 5J</p>
<p>Work done on box = work done by man + work done by friction = 10 - 5 = 5J</p>
<p>does Work=F * displacement or distance
say the object is moved 10 meters along a curvy path. But its displacement is only 5 meters. Which value do you use as d?</p>
<p>It depends on whether the acting force is a conservative force or a non-conservative force.</p>
<p>Friction is non-conservative, so you will use distance. Gravitation force is conservative, so you will use displacement.</p>
<p>The formula W=F*displacement (here * is the dot product and F and displacement are vectors) does <em>not</em> require the force to be conservative. However, it only applies in special cases where the direction and magnitude of the force F are constant. For a curvy path (like pushing a box along a floor but not in a straight line), the friction force changes direction along the path so the formula doesn’t apply.</p>
<p>Luckily, for typical friction cases, the friction force magnitude doesn’t change and its direction is opposite to the instantaneous direction of motion, so to get the work done by friction you can use W = -F<em>(path length) = -F</em>distance (here, * is regular multiply).</p>