1 Physics Q.

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1

<p>Hey. Do you mind helping me out with this physics problem?</p>

<li>A 74 kg man drops to a concrete patio from a window only 0.50 m above the patio. He neglects to bend his knees on landing, taking 2.0 cm to stop.
(a) What is his average acceleration from when his feet first touch the patio to when he stops?</li>
</ol>

<p>(b) What is the magnitude of the average stopping force?</p>

<p>I’m not really sure where/how to begin this problem, but I started calculating the time that it took him to fall:</p>

<p>x-x0 = v0t + 1/2 at^2
x-x0 = 1/2 at^2
t = sqrt(2(x-x0)/a) = sqrt(2(0.50)/9.8) = 0.319438282 s.</p>

<p>Then I went into a different direction. I looked at when the man is touching the patio for the first time:</p>

<p>^ F = ?
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v mg = (74)(-9.8) = 725.2 N.</p>

<p>F = ma
F - 725.2 = (74)(a)</p>

<p>Now I’m stuck, because I don’t know F nor a. Can anyone help me out?</p>

2

<p>Ummm. First find the velocity when he reaches the ground, then use vf^2=vi^2-2ad to find acceleration. Use 0 for vf, whatever velocity you found for vi, and 2cm for d</p>

3

<p>Note that the velocity when he hits the ground is not 0 at this point. Only after he travels an extra 2 cm does the deceleration take place and he finally stops</p>

4

<p>Something didn't come out right. Here is what I did:</p>

<p>v^2 = v0^2 + 2ad
v^2 = 2ad
v = sqrt(2ad) = sqrt(2<em>9.8</em>0.50) = 3.130495 m/s.</p>

<p>v^2 = v0^2 + 2ad
0 = (3.130495)^2 + 2a(0.02)
a = 244.9999736 m/s^2.</p>

<p>The acceleration is a little bit too big...don't you think?</p>

5

<p>it is supposed to be too big, since he goes from some velocity to 0 in such a short distance</p>

6

<p>For part (b), it says to find the magnitude of the average stopping force. Why can't I just do this:</p>

<p>F = ma = (74)(245) = 18130 N?</p>

7

<p>That's how you do it</p>

8

<p>When I enter that into WebAssign (program my school uses to do homework), it says it's wrong.</p>

9

<p>Are you on impulse-momentum yet? If so, try F(time) = m (change in velocity), you have to calculate the time though</p>

10

<p>you forgot to include the force of gravity with your final acceleration calculation. You should be using Aaverage=m(9.8+a)</p>

11

<p>Do you mean Average Stopping Force = m(9.8+a)?</p>

<p>Should I add 9.8 to acceleration then multiply by mass, or should I subtract 9.8 from acceleration?</p>

12

<p>I'm not really sure which way my acceleration is going, since it's the average. I only know that 9.8 is going down. I'm not sure if the mans average accereration is going in the same direction or opposite that of gravity.</p>

13

<p>"prelic" is right. If you think about the situation, a dude jumps out of a window and takes him to 2cm to stop. As soon as he touches the patio floor, there is that big accelerating force against him to stop. Therefore, if he's being stopped, the acceleration has to go in the opposite of gravity, otherwise he'd fall through the Earth (which would be very odd). So yeah, it's f=m(g+a).</p>

<p>This is also seen when someone is riding in an elevator. You feel "heavier" when the elevator is going up because there's an acceleration going up in addition to gravity pulling down. However, if the elevator was going down, you'd feel "lighter", which is why you add when going against, and subtract when going with gravity.</p>

<p>Does that make sense or am I completely wrong?</p>

14

<p>I can't seem to figure out what I did wrong for this problem.</p>

<p>Two blocks (m = 16 kg and M = 88 kg) are not attached to each other. The coefficient of static friction between the blocks is mu_s = 0.38, but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force F required to keep the smaller block from slipping down the larger block?</p>

<p>Here's what I have so far after 2 days of working through it:</p>

<p>My FBD looks like this (in words):
F_fr = up; mg = down; N = left; F = right for the small block.</p>

<p>F<em>fr - mg = 0
F - N = 0
mu</em>s *N = mg
mu<em>s *F=mg
F = mg/mu</em>s</p>

<p>Plugging in numbers to this equation yields a minimum force of 413 N. The correct answer is actually 490 N.</p>

<p>Do I need to take into account the FBD for the larger block? I'm unsure of the forces on the horizontal direction. I know that N is up and mg is down for the larger block only.</p>

15

<p>Are these blocks aligned at an angle? Is the smaller block on top of the larger block?
Never mind, it seems as if the smaller block is being pressed against the larger block by a horizontal force and being held by that force alone, is that it?
Is the larger block pressed against a wall or something similar? Because you might have forgotten to account for the fact that both blocks might be accelerating in the horizontal direction</p>

16

<p>The blocks are arranged like a "7" shape. The small block is on the top portion of the side of the large block. There is no wall. We don't know if it is accelerating or not (I'm assuming it's not since the mass of the large block is so much greater than the small block). The blocks are not at an angle.</p>

17

<p>Well you said the large block and the floor are frictionless so they must be accelerating since there is no opposing frictional force between the block and floor.</p>

18

<p>Ok. I redid the problem with acceleration now.</p>

<p>Now I get F = (mg + mu<em>s *ma)/mu</em>s. How do I get the acceleration a?</p>

19

<p>I think I got it. To get the acceleration a, I looked at the entire system consisting of both masses.</p>

<p>F = (m + M)a
a = F/(m+M)</p>

<p>I plugged a back into F = (mg + mu<em>s *ma)/mu</em>s and solved for F. Now I get 487.65 N, which is much closer to the correct answer of 490 N.</p>

20

<p>That looks good. The discrepancy is probably due to sig figs</p>