<p>Right, right. Oops.</p>
<p>B (3 sec) is right.</p>
<p>You guys are too fast. Here is a very challenging one.</p>
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</tr>
</thead><tbody>
<tr>
<td>—a-----</td>
</tr>
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<td>---------</td>
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<td>---------</td>
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<p>=======================table================</p>
<p>Block a (16 kg) is moving on a frictionless table, being pushed by a force F on the left. On the right of block a is block b (4 kg), which is being held in place by friction between a and b (coeff = 0.5).</p>
<p>The smallest possible value of F is:</p>
<p>A) 50 N
B) 100 N
C) 200 N
D) 400 N
E) 800 N</p>
<p>is it B?</p>
<p>Block A should move to the right until it sticks to B and moves together with it. So the force being applied should be equal to or more than the friction force of Block A and B. </p>
<p>F(friction)=coefficient x normal force (which in this case is mg)</p>
<p>mg=20 X 10</p>
<p>so F(friction)=100</p>
<p>So, the force should be more than or equal to 100…</p>
<p>Or so i think…</p>
<p>yeah but i have one problem withur approach. i dont understand how one can use the coefficient of friction between a and b the way u do.</p>
<p>fig answer please</p>
<p>is B not touching the table then? </p>
<p>does the question mean that the friction between a and b is preventing b from falling to the table?</p>
<p>in that case…</p>
<p>since the friction force is acting upward, against gravity</p>
<p>coefficient X normal force=16g+4g=200</p>
<p>normal force=400</p>
<p>normal force=force acting left</p>
<p>so ans would be 400N…</p>
<p>answer please!!!</p>
<p>Block b is not touching the table, it is held there hanging due to the friction between block a and b. Guess my diagram wasn’t so good :(</p>
<p>Hint: block b has a net vertical force of zero</p>
<p>so is 400N right?</p>
<p>i mean… the gravity pulling the block B down meets a resisting force, friction, thus cancelling each other out. </p>
<p>But, as to the magnitude of mg, is it 4g (just the block B) or 20g (gravity pulling both a AND b??)</p>
<p>im rele confused…</p>
<p>correct me if im wrong</p>
<p>but 4g=coeff X normal force</p>
<p>normal force=80N</p>
<p>so should 80N be in one of the ans choices?</p>
<p>So the friction force between blocks a and b need to cancel the weight of block b in order to keep it there. So,</p>
<p>u(Fnormal)=mg
.5(Fnormal)=(4kg)(10)
Fnormal=80N</p>
<p>So I guess the answer is B, 100N? 50N would be too small to keep block b connected to block A right?</p>
<p>Answers please =] I don’t really know if I’m right.</p>
<p>^
Yeah…
That’s what I got, too!</p>
<p>Hm ok, this is what I think:</p>
<p>The acceleration of the system is a= F / (16 + 4) = F/20.</p>
<p>Just considering the 4 kg mass, we have the system:</p>
<p>N= ma = (4)(F/20) = F/5.</p>
<p>We also know that (mu)N = mg. So, we have</p>
<p>(.5)(F/5) = (4)(10)
F = 400 N.</p>
<p>(D).</p>
<p>So far in practice tests i have gotten all 800s in Kaplan, and a 48 raw score in Barron’s. What would you predict I’d get on the real thing tomorrow?</p>
<p>Two blocks answer: D ( 400 N ) … nice job, Rupac, I made that one very hard.</p>
<p>are the questions that you are asking even realistic??? they seem far too convoluted for the sat physics to me</p>
<p>Yeah there is no way that block problem will be on there tomorrow.</p>
<p>Probably some are too easy, and others (like that last one) are too hard for the level of the subject test. But even too hard questions can be useful if you give them a try.</p>
<p>Here’s another (easier) one.</p>
<p>Three 1 ohm resistors are arranged in a right triangle with one resistor on each side of the triangle. A 1 volt battery is attached, positive side on one end of the hypotenuse, and negative side on the other end. The current through the battery is:</p>
<p>A) 1/2 A
B) 1 A
C) 3/2 A
D) 2/3 A
E) 1/3 A</p>
<p>Is the answer choice E?</p>
<p>Hm, I got (C).</p>
<p>The node with two one-ohm resistors and the node with one resistor are in parallel. So we have a net resistance of:</p>
<p>1/(1/1 + 1/2) = 1/(3/2) = 2/3 ohm.</p>
<p>By Ohm’s Law,</p>
<p>V = IR
1 V = (I)(2/3 ohm)
I = 3/2 A.</p>
<p>bumping up</p>