Kinetic friction, mass on track when will it stop.

  • Thread starter sg001
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  • #1
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Homework Statement



A block of mass m is initially held in place on the left side of a track at a distance h above the bottom of the track. The track is completely frictionless, with the exception of a rough horizontal section of track of length h which has a coefficient of kinetic friction μ = 0.15. If the block is released from rest, how many times does the block completely cross the rough section of track before it stops?




Homework Equations






The Attempt at a Solution



So I have worked out the height to be 0.85 m, but am stuck on finding an expression for when the ball will stop I have what I think it is

x=number of times without stopping= v^2/2gh

does this see right, and I worked out v = 11.43 but when I plug it in I get 7.84 times... in the answer booklet it says seven but im not sure if thats because you cant round up because it doesnt make sense so they round down to 7??
 

Answers and Replies

  • #2
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Hey,


I hope this makes it clear enough :-)


how many times does the block completely cross the rough section of track before it stops?
 
  • #3
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Wondering how you get the value of h=0.85 m since only a coefficient of kinetic friction μ = 0.15 is given. And not mentioning anything about a ball.
 
  • #4
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Wondering how you get the value of h=0.85 m since only a coefficient of kinetic friction μ = 0.15 is given. And not mentioning anything about a ball.
by the work energy theorem, I worked out how much energy was lost through the distance of friction patch... the height of the track is the same 'height' or distance as the friction patch.
 
  • #5
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Hey,


I hope this makes it clear enough :-)
Yeah thanks, I was assuming that just wanted to check... I did not really pick up on the COMPLETELY part of the question.
 

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