Distance spring decompresses when friction is involved


by becky_marie11
Tags: force, friction, kinetic energy, spring, work
becky_marie11
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#1
Feb9-12, 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
vo=8.50m/s
k=68.00N/m
μ=.200
x=?


2. Relevant equations
Kf-Ki=W
K=1/2mv2
Fs=-kx
Ffric=μNf
W=Fd



3. The attempt at a solution
First I found the work of the whole system (friction and spring forces) by using
Kf-Ki=W.
For the initial kinetic energy, I used
K=1/2mv2=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+μNf)x
I work it out to a quadratic by plugging the numbers in...
-68x2+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!!
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Doc Al
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#2
Feb9-12, 02:59 PM
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Quote Quote by becky_marie11 View Post
Then, using
W=Fd (just multiplying the forcexdistance because both vectors are in the same direction)
Careful: That only works when the force is constant. The spring force is not constant.

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.
tiny-tim
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#3
Feb9-12, 03:03 PM
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hi becky! welcome to pf!
isn't it 1/2 kx2 ?

xlava
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#4
Feb9-12, 03:20 PM
P: 33

Distance spring decompresses when friction is involved


Quote Quote by tiny-tim View Post
hi becky! welcome to pf!
isn't it 1/2 kx2 ?
W = ∫Fdx (F dot dx, technically)
Fspring = kx

W = ∫kxdx
Wspring = kx2/2

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.
becky_marie11
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#5
Feb12-12, 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?
tiny-tim
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#6
Feb12-12, 02:49 AM
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hi becky!
Quote Quote by becky_marie11 View Post
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?
yes!!

work done is force "dot" distance, and 1/2 kx2 is 1/2 kx (average force) times x (distance)!
becky_marie11
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#7
Feb13-12, 05:25 PM
P: 8
Thank you!!!


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