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Energy loss due to friction 
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#1
Feb2211, 04:51 PM

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1. The problem statement, all variables and given/known data
A 15.0 kg block is dragged over a rough, horizontal surface by a 74.9 N force acting at 17.6 degrees above the horizontal. The block is displaced 4.91 m, and the coefficient of kinetic friction is 0.266. Find the work done by the 74.9 N force = 351 J Find the work done by the normal force = 0 J What work does the gravitational force do on the block? = 0 J This is the part i can't figure out: How much energy is lost due to friction? And related to that: Find the total change in the block's kinetic energy. 2. Relevant equations Ff = ((mu)k)(mg) E = F*d 3. The attempt at a solution Ef = ((mu)k)(mg) (d) = 0.266*15*9.8*4.91 = 192 J What am I doing wrong? 


#2
Feb2211, 07:36 PM

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The friction force is not (mu_k)mg, it is (mu_k)N, where N is the Normal force, which , in this problem, is not equal to mg. Solve for the normal force first. The use the work energy equation or other method to get the change in KE.



#3
Feb2211, 07:41 PM

P: 28

So it'll be : Ff = (0.266)(74.9 sin17.6)
= 6.02 N 


#4
Feb2211, 07:47 PM

P: 28

Energy loss due to friction
Then I took that and put it in to W = Fd = (6.02)(4.91) = 29.6 J
Still wrong...? 


#5
Feb2211, 07:47 PM

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Draw a Free Body Diagram. 


#6
Feb2211, 07:49 PM

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Huh? do i use the mass in there somewhere?



#7
Feb2211, 08:45 PM

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When you draw a free body diagram, you note all forces acting on the block , in both the x and y directions. There are three forces acting in the y direction, one of which is the component of the applied force which you have correctly calculated. What are the other 2 forces acting on the block in the y direction? Then use one of newton's laws to find the unknown force in that direction.



#8
Feb2411, 08:46 AM

P: 28

So the frictional force is:
Ff = (mu)k * N = (0.266)* (74.9sin17.6 + (15*9.8)) = 45.1 N But how would I calculate the work done? The block doesn't move in the ydirection, so I would think no work coule be done since W=Fd 


#9
Feb2411, 09:00 AM

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#10
Feb2411, 09:40 AM

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Okay, so my normal force would then be
Fg = N + vertF N= Fg  vertF = (15*9.8)  (74.9sin17.6) = 124 N then my frictional force would be: Ff = (mu)k * N = (0.266) * (124N) = 33.0 N Work done by friction: W = Fd = 33N * 4.91m = 162 J 


#11
Feb2411, 11:28 AM

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