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Springs Help Needed Plz

  • Thread starter brunie
  • Start date
  • #1
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Help concerning springs.

A 2.00 kg block is attached to a spring of force constant 560 N/m as in Figure. The block is pulled 5.95 cm to the right of equilibrium and released from rest.

(a) Find the speed of the block as it passes through equilibrium if the horizontal surface is frictionless.
(b) Find the speed of the block as it passes through equilibrium (for the first time) if the coefficient of friction between block and surface is 0.350.

Ee = 0.5kx^2
F = -kx
Ek = 0.5mv^2
Ff = umg

A)
So Ee = 0.5 (560 N/m) (0.0595 m)^2
= 0.99127

Ee gets tranferred into Ek

So 0.99127 = 0.5 mv^2
1.98254 = (2 kg) v^2
v = 0.9956

B)
Ff = umg
= (0.350)(2 kg)(9.8)
= 6.86

I don't kno how to relate the frictional force into energy. I know that the block should be slower than in part A) tho.
 

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Answers and Replies

  • #2
Doc Al
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44,892
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I don't kno how to relate the frictional force into energy.
Consider the work that must be done against friction--that will tell you the amount of mechanical energy that is dissipated as "heat".
 
  • #3
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Assuming that this is an intro course and ordinary differential eqns are not to be used, which would lead to a soln w/o using energy,
frictional work=integral(ff)dx where ff is as you state. Multiplying that by 0.0595M should be the energy lost to friction. So you have the potential energy of the spring less the work done by friction equaling Ke as it passes thru equil.
 
  • #4
62
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ok so
Ek = Ee - (0.0595 m)(6.86 N)
Ek = 0.99127 - 0.40817
0.5 (2 kg) v^2 = 0.5831
v = 0.7636 m/s

would this be right?

thanks in advance to all those who helped
 
  • #5
Doc Al
Mentor
44,892
1,143
I haven't checked your arithmetic, but your method is exactly correct.
 

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