Force of friction quick question

  • Thread starter Tycho
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  • #1
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a block of mass 10kg is attached to a spring with a constant of 1850 N/m. the coefficient of sliding friction between the block and the surface is .25
the block is pulled back 11 cm to the right and released.

the question asks the speed at different points, and how far the distance after equilibrium it travels.

THe problem i have is this.... i am fine on all of this except for how to figure the friction into the equation? i know the velocity neglecting friction, and i know the force of friction acting on the box. i need something to put the two together. can someone please lend me a small hand?
 

Answers and Replies

  • #2
Doc Al
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The stretched spring has a certain PE. If there were no friction, as the spring contracted [itex]-\Delta \texrm(PE) = \Delta \texrm(KE)[/itex]. But with friction, some of that PE is transformed to heat via work done against friction. So: [itex]-\Delta \texrm(PE) = \Delta \texrm(KE) + W_\texrm{(friction)}[/itex]. What's the work done against friction?
 
  • #3
jamesrc
Science Advisor
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How about considering the energy?

[tex]\Delta K + \Delta U = -fd [/tex]

where f is the friction force, d is the distance travelled, and K and U are kinetic and potential energy, respectively. This is a statement that says the change in total energy of the system is equal to the work done on the system (the work done on the system in this case is negative (frictional loss)).

Try take it from there; we can help more if you need it.
 

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