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Block on Floor with Friction 
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#1
Oct608, 10:04 PM

#2
Oct608, 10:16 PM

P: 328

well energy is always conserved. Therefore: [tex] \frac{1}{2}mv_1^2 + F\Delta d = \frac{1}{2}mv_2^2 + F\Delta d [/tex]
So you know that Friction is playing no part in the initial velocity so you can get rid of Fd from the LHS. You cancel the masses, you have initial velocity, distance and the retarding force of friction. Solve for v_{2} 


#3
Oct608, 10:26 PM

P: 43

That's exactly it. I got the right answer. I was setting it up like this:
(1/2)mv(final)^2  (1/2)mv(initial)^2 = Ffriction over the distance. My problem with this set up was i wasn't making the friction negative. That's why i was getting the wrong answer. Thanks so much for your help! 


#4
Oct3108, 06:36 PM

P: 1

Block on Floor with Friction
I'm still stuck on this problem... Using this equation: (1/2)mvf^2(1/2)mvi^2= (F*d) or coefficient of friction*mgd
So my final equation to solve for vf is: vf= (squareroot): 2(coefficient of friction*gd+vi^2) But my answer is still wrong. Could someone please help me? It must be the numbers I'm plugging into the equation. Here are the number given: A box slides across a frictionless floor with an initial speed v = 2.8 m/s. It encounters a rough region where the coefficient of friction is ľk = 0.7. (b)If instead the strip is only 0.17 m long, with what speed does the box leave the strip? I found that the shortest length of rough floor which will stop the box is 0.57m in part (a) 


#5
Oct3108, 06:50 PM

P: 328

[tex] \frac{1}{2}mv_1^2 + F_1\Delta d_1 = \frac{1}{2}mv_2^2 + F_2\Delta d_2 [/tex] because there is no initial friction, we can get rid of the [tex] + F\Delta d [/tex] on the left hand side, or get rid of the one on the right hand side and make the one on the left hand side negative (both yield the same answer mathematically). solve for v_{2} 


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