<p>F isn’t mg…</p>
<p>F is the force necessary to compress the spring by the .020 meters (x). If F were mg, then k would vary, which goes against the definition of the k value (constant for a particular spring).</p>
<p>anyhoo. i think i got full credit for all except for #3 part B, I forgot the magnetic force, so my tension force will be off ;X</p>
<p>I wasn’t sure on that one, but isn’t only the frictional force opposing the tension force?</p>
<p>nop, friction and the magnetic force, if you forget to do friction, when u compare the work done to the energy of the resistor you will get that theyre equal which is wrong, the work is greater than the energy</p>
<p>doesn’t magnetic force point to the top of the page though. (perpendicular to the tension force)</p>
<p>guys there was NO magnetic force in that problem wth
there was electromagnetic induction, which was the point for that problem…
E=BlV</p>
<p>it’s not like a point charge where F=qvB or a wire where F=IlB </p>
<p>the only force opposing the movement of the rod was the frictional force</p>
<p>the magnetic force points to the left (F=BIlsinx), so right hand rule it and you’ll see that it points to the left</p>
<p>V=Blv is derived from the magnetic forces acting on the rod, so you’re making it too complicated!!</p>
<p>the force required for constant velocity of the rod
is F=ukFn</p>
<p>the rod acts as a wire pointing up</p>
<p>oh wait nvm i see
darn i missed it!!</p>
<p>there was electromagnetic induction, but lenz’ law created a force to the left.
so you had Fb and friction going left, tension going right. lol</p>
<p>yea, that’s it even purposefully says the rod has no internal resistance and neither do the rails lol, but don’t worry you should still get full credit for the part about work as long as u used w=fd and the calculations were made right using the wrong force.</p>
<p>So can someone write the answers for that question then?
And write steps? please?!</p>
<p>=[=[
crap crap crap
ugh!!! stupid me
what was your answer for part c and d for that problem???</p>
<p>part d won’t matter if you got part b wrong, then part d is just b*2, lol</p>
<p>c idk, part a^2<em>3</em>2</p>
<p>okkkk
a) you have Emf=IR and Emf=bLv.</p>
<p>Set them equal to each other and solve for I-the current. So I=bLv/R
You have b-the magnetic field, L-length, and v-velocity, and R-resitance</p>
<p>b)Tension=Magnetic Force+Friction.
Friction=ukFn, where Fn=mg.</p>
<p>c)P=IV or in this case P=I^2<em>R, then to find the energy used by the resistor in 2s, find work. so P=W/T so W=P</em>T.</p>
<p>d)W=FD, where F is the tension calculated in part b, and you find distance by kinematics where x=v*t since there is no accel.and t is 2s.
e) the work done by the tension is greater than the energy dissipated by the resistor b.c it takes more energy moving the bar due to the force of tension which is equal to magnetic and friction than the energy by the resistor which depends on the magnetic field only (i guessed this part so idk)</p>
<p>If the bar is moving to the right and the magnetic field points into the page, doesn’t the right hand rule say that the magnetic force would point up?</p>
<p>no, right hand rule says nothing about the BAR. there’s moving charges moving UP the bar, so right hand rule says force is <</p>
<p>How do you know that the charges are moving up?</p>
<p>Also, if I did everything correctly in question 3 but forgot about the friction force, what do you think will be my score on that question?</p>