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Horizontal Spring-Mass System with Friction

  • Thread starter Tzvika
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I finished with my first semester of physics in Spring, and I thought that I had understood the unit on springs pretty well; however, today a friend gave me this problem to solve, and I’m drawing such a massive blank, it’s not even funny. I’m so frustrated with this problem.

Homework Statement



The diagram is drawn as follows: there are five masses (from left to right: M1, M2, M3, M4, M5) connected to each other by springs, resting on a horizontal surface. M1 = 2kg, M2 = 2kg, M3 = 3kg, M4 = 3kg, M5 = 5kg. Between M1 & M2, the speed of the force is 100 N/m; between M2 & M3, the speed of the force is 200 N/m; between M3 & M4, the speed is 200 N/m; between M4 & M5, the speed is 200 N/m. The force is being applied to M5 [furthest right], direction left.

The problem:
An external force of magnitude 90 N is applied to a system of blocks connected by springs. After some time the system will move as a whole with constant acceleration. Find this acceleration. The coefficient of kinetic friction between the surface & the blocks is mu-kinetic = 0.1...

Homework Equations



Fnet = ma = 0
Ffriction = mu*Fnormal = mu*abs(Fgrav)
Fspring = -kx, where x = displacement of spring from equilibrium

The Attempt at a Solution



What does N/m mean with regards to springs? Is that how much force is transferred to the block next in line in the spring-mass system? Am I to use that relationship to find out what x is? But I don’t even know k. Do I? I’m setting the origin at M5, with the left direction as negative.

If I’m pushing on M5, the horizontal net force is Fnet-horizontal = -90 N + Ffriction = -90 N + mu*abs(Fgravity) = -90 N + 0.1*abs(5*-9.8) = -85.1 N

So then what? Substitute -85.1 N in to the relationship 200 N/m to solve for m? If I do, I end up with -0.4225 m. Then, using -85.1 N and -0.4225 m, do I substitute that in to find the value of the spring constant? Because it doesn’t look right at all. I come up with k being 201.4 ...

But why am I even finding the spring constant? I feel like I’m sort of really lost, and like I don’t understand springs at all. @_@;; I looked through my book, and again, massive blank.
 

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Answers and Replies

  • #2
gneill
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What do you think is meant by the statement, "After some time the system will move as a whole with constant acceleration"? Why "after some time"? What might happen during that time?
 
  • #3
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Wow, I’m an idiot. a isn’t zero. Constant.

Ok, so before that point in time when a becomes constant and the blocks are changing speed at a constant rate, the masses are changing speed at different rates from one another, meaning the horizontal force on each block is different?

It would take some time because the blocks are separate and would not immediately feel the entirety of the force?
 
  • #4
gneill
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Wow, I’m an idiot. a isn’t zero. Constant.

Ok, so before that point in time when a becomes constant and the blocks are changing speed at a constant rate, the masses are changing speed at different rates from one another, meaning the horizontal force on each block is different?

It would take some time because the blocks are separate and would not immediately feel the entirety of the force?
That's more or less it. When the force is first applied there will be an impulse that will travel through the spring/mass system, setting up various oscillations. But note that there is friction involved (damping). So after some time you can expect the oscillations to die down, and the system will settle into a steady state with each spring compressed according to its spring constant and the net force acting on it. How would you determine the acceleration of the ensemble after that time?
 

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