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Distance spring decompresses when friction is involved 
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#1
Feb912, 02:52 PM

P: 8

1. The problem statement, all variables and given/known data
A block of mass m = 4.30 kg slides along a horizontal table with speed v0 = 8.50 m/s. At x = 0 it hits a spring with spring constant k = 68.00 N/m and it also begins to experience a friction force. The coefficient of friction is given by μ = 0.200. How far did the spring compress when the block first momentarily comes to rest? m=4.30 kg v_{o}=8.50m/s k=68.00N/m μ=.200 x=? 2. Relevant equations K_{f}K_{i}=W K=1/2mv^{2} F_{s}=kx F_{fric}=μN_{f} W=Fd 3. The attempt at a solution First I found the work of the whole system (friction and spring forces) by using K_{f}K_{i}=W. For the initial kinetic energy, I used K=1/2mv^{2}=155J. For the final kinetic energy, I used 0, because I'm measuring the distance after the mass came to a stop. So, plugging the numbers into the Work equation, I simply get W=155J I'm pretty sure I'm on the right track so far? Then, using W=Fd (just multiplying the forcexdistance because both vectors are in the same direction) I plugged in: W=(kx+μN_{f})x I work it out to a quadratic by plugging the numbers in... 68x^{2}+8.44366x+155=0 I solve then for x (using the quadratic root finder on my calculator, so I'm not screwing up the quadratic..), but both of the answers it gives me are incorrect.. What am I doing wrong? Help please!! 


#2
Feb912, 02:59 PM

Mentor
P: 41,293

Hint: What's the energy stored in a compressed spring? Also: Careful with signs. The work done on the mass by both the spring and the friction force is negative. 


#3
Feb912, 03:03 PM

Sci Advisor
HW Helper
Thanks
P: 26,160

hi becky! welcome to pf!
isn't it 1/2 kx^{2} ? 


#4
Feb912, 03:20 PM

P: 33

Distance spring decompresses when friction is involved
Fspring = kx W = ∫kxdx Wspring = kx^{2}/2 Just to show where it came from, since W does NOT equal Fx, that is not a definition, it is a derivation of work for a constant force. 


#5
Feb1212, 12:23 AM

P: 8

Thank you guys! I finally figured out the right answer! But when I solve for the quadratic equation, why is the distance vector only distributed to the friction force? This is what I tried to do...
W=(1/2kx^2+friction force)*x and this is what is right... W=1/2kx^2+friction force*x Is it because the work of the spring is already in terms of "work" and the friction force isn't? 


#6
Feb1212, 02:49 AM

Sci Advisor
HW Helper
Thanks
P: 26,160

hi becky!
work done is force "dot" distance, and 1/2 kx^{2} is 1/2 kx (average force) times x (distance)! 


#7
Feb1312, 05:25 PM

P: 8

Thank you!!!



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