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krystal38garcia87

  • 2 years ago

A bead slides without friction around a loopthe-loop. The bead is released from a height 20.3 m from the bottom of the loop-the-loop which has a radius 7 m. The acceleration of gravity is 9.8 m/s^2. How large is the normal force on it at point A if its mass is 3 g?

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  1. krystal38garcia87
    • 2 years ago
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    I already calculated speed. 11.112 m/s. To find the answer to this problem do I use the formula. F=mv^2/R = ((.005kg)(11.112 m/s)^2)/5m = 0.123N ?

  2. JamesJ
    • 2 years ago
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    A is the top of the loop?

  3. hosein
    • 2 years ago
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    if point "A" in top normal force is N=(mv^2/r)-mg if "A" in bottom N equal to:|dw:1329244817817:dw|(mv^2/r)+mg

  4. hosein
    • 2 years ago
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    hence in bottom 2nd law of newton is N-mg=mv^2/r & on top is N+mg=mv^2/r

  5. krystal38garcia87
    • 2 years ago
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    top... thainks guys.

  6. krystal38garcia87
    • 2 years ago
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    .02347N ?

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