Problem with spring + inclined plane + conversation of energy etc

  • Thread starter -sandro-
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  • #26
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At point D, you have KE and PE, so at point E, the KE will be increased by the change in PE from D to E. Now that you have the KE at E you can find the velocity at E.

Just before E, the weight will be the sum of the centripital force as determined by the velocity you calculated above + mg. Just after E, the weight will be only mg. Now kowing the velocity, you can probably fined the distance to stop.

Did you get the velocity at D?
 
  • #27
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I didn't can you please tell me what equation you used to find the time to fall to point D?

For the change in ME from D to E seems like you're confirming what I already said, good!:D
 
  • #28
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OK, so think about this. If there wassn't any air drag then you can easily find the velocity at D given that the change in PE from C to D is converted into KE. Now with air drag, the velocity at E will be a little slower. If you are into doing differential equations, then you can set up the problem and solve. However there is an already computed formula that works when the drag is proportional to V as opposed to V^2. In this case the Drag is proportional to V.

The equation is.. V =(mg/b)(1-e^(-tb/m)) perhaps you have seen this equation. Notice that as t becomes very large, V becomes the terminal velocity of mg/b.

You don't have t however so if you integrate v(t) you get S(t). So first knowing S = 2 m, you can find t ujsing the integration you just did. Then knowing t you can find v. When you integrate v(t) to get S(t) don't forget to find the constant of integration using the fact that at t = 0, S(t) = 0.
 

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