<p>Hi, I had a specific question about physics:</p>
<p>Work = Force (displacement)</p>
<p>so would exerting force to move something one full loop around a track do work? I feel like the answer is yes, despite being contrary to work's definition.</p>
<p>good luck to everyone taking SAT II's tomorrow! I'm taking Math IIC and Physics</p>
<p>yes and it isn't contrary to work's definition</p>
<p>but don't you end up in the same place, ie displacement = 0?</p>
<p>so what? Going around a loop in a conservative field = no work such as
q(int(E.dl)).</p>
<p>q(int(E.dl? I am sincerely confused by your short explanations. Elaborate more clearly, please.</p>
<p>Work can be zero even when there's a non-zero force. Work is path-dependent. This article may help: Mechanical</a> work - Wikipedia, the free encyclopedia .</p>
<p>In a conservative electrostatic field, if you go around a closed path regardless of the shape, the sum of the component of the force = q(charge) * E (electric field) is zero because the infinitesimal sums add up to zero.</p>
<p>If a car goes around a track, is work done?? Isn't energy expended?</p>
<p>Is the physics SAT II calculus based like the AP test?</p>
<p>Physics B is not calculus based, and the SAT II one isn't either. You don't even get a calculator for the phys sat II</p>
<p>thanks, so work is NOT done moving one loop around a circular track b/c displacement = 0. k</p>
<p>Work is done because the article neglects friction which is not in the same direction as the centripedal force. In an ideal frictonless system this is true for mechanics type problems. If charge is moved around a time dependent electric field (the force is time dependent) work is involved (Faraday's law).</p>
<p>thanks doc!</p>
<p>I thought it was Work = Force*Distance.</p>
<p>I don't understand your question?</p>
<p>k guys hopefully this will clear some things up. Work = Distance travelled (or displacement or whatever you want to call it) * The component of the force in the same direction as the distance traveled. So if you have a 10kg block with 3N applied at 30 degrees to the horizontal to push it 10 meters, the work done will be, neglecting friction, 3N<em>cos30degrees</em>10m. The normal force in this situation, which is roughly 100N does no work because it is perpendicular to the direction of travel. 100N<em>cos90degrees</em>10meters = 0.</p>
<p>Now lets consider this car going around the track situation again. The only force on the car, or particle, or whatever happens to be moving in the circular motion is centripetal force. This force is ALWAYS perpendicular to the direction of travel. That is how the object moves in a circle... from what i just explained it should be clear that no work is done, not only because it has no displacement after one loop, but also because the centripetal force was perpendicular to the direction of motion.</p>
<p>Keep the physics questions coming. SAT2 tomorrow for me too =)</p>
<p>Nope, displacement. </p>
<p>"the mechanical work can be calculated from the scalar multiplication of the applied force (F) and the displacement (d) of the object. "
Mechanical</a> work - Wikipedia, the free encyclopedia</p>
<p>Work is the integral of the scalar product of F and dl where dl is an infinitesimal displacement or distance moved. F, the force can be due to a vector field that varies as a function of position and time. That is why the expression is integrated.</p>
<p>Here is a physics question for all you smart folks. I'm assuming e&m is on this test? Two negative charges repell each other. Two parallel wires with current going in the same direction attract each other. Yet if one walks with the moving charge in the wires, the charges appear stationary and should repell each other - explain why if you are stationary with respect to the wire, they would attract while if you move with the charge they should repell.</p>
<p>Here's an easy example of W > 0 but total displacement = 0. You push a crate with a constant force (in magnitude) around a circle on the floor, coming back to the original spot. The direction of your force is always tangential to the circle. In this case, the force is always in the direction of motion and W > 0.</p>