# Block on Floor with Friction

by Awwnutz
Tags: block, floor, friction
 P: 43 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 µk = 0.7. a) What is the shortest length of rough floor which will stop the box? b) If instead the strip is only 0.28 m long, with what speed does the box leave the strip? Relevant Equations: Work Energy Theorem, Chane in Kinetic Energy = Work done on box. I already found the answer to part a.) (1/2)mv(final)^2 - (1/2)mv(initial)^2 = Ffriction where i got: (1/2)(2.8)^2 = 0.7(9.81) * x when i solved for x i got 0.571m But its part b.) that i'm having the problem with. I set the problem up the same but solved for the velocity while i put in .28m for the distance 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
 P: 328 well energy is always conserved. Therefore: $$\frac{1}{2}mv_1^2 + F\Delta d = \frac{1}{2}mv_2^2 + F\Delta d$$ 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 v2
 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!
 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 µk = 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)
P: 328
 Quote by manleju 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 µk = 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)
So, we know that energy is going to be conserved. First I'd write out a balanced equation using all the types of energy we have in this question:

$$\frac{1}{2}mv_1^2 + F_1\Delta d_1 = \frac{1}{2}mv_2^2 + F_2\Delta d_2$$

because there is no initial friction, we can get rid of the $$+ F\Delta d$$ 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 v2

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