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catsrule332

  • 4 years ago

A 25 kg block is at rest on a horizontal surface. A horizontal force of 75 N is required to set the block in motion. When it is in motion a 60 N forced is required to keep the block moving with a constant speed. Find the coefficients of static and kinetic friction. Draw a force diagram and explain!

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  1. salini
    • 4 years ago
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    so catsrule wat have u figured out so far in this question?

  2. catsrule332
    • 4 years ago
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    find normal force: n=mg=(25)(9.8)= 245 N \[f=m*\eta\]|dw:1333113193720:dw|

  3. JamesJ
    • 4 years ago
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    When the object is not moving, the maximum frictional force on the block is the static friction. Static friction is given by \[ Fr_S = \mu_S N \] where \( \mu_S \) is the coefficient of static friction. Now which force in the problem is that equal to?

  4. salini
    • 4 years ago
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    u are in the right track wat is the expression for frictional force with normal reaction? and do u know the difference abt co-efficient of static friction and kinetic friction?

  5. JamesJ
    • 4 years ago
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    On the other hand, once the object is moving, the maximum frictional force is the kinetic friction and that is given by \[ Fr_K = \mu_K N \] where \( \mu_K \) is the coefficient of kinetic friction and of course \( N \) is the normal force. What value is the kinetic friction equal to in this problem?

  6. salini
    • 4 years ago
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    cats rule james has given everything needed for the sum u have to get the answer using that

  7. catsrule332
    • 4 years ago
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    Static friction: fsmax=\[\mu *\eta\] Kinetic friction: fk=\[\mu * \eta\] 75/(25)(9.8)= .31 N 60/245) =.24 N

  8. JamesJ
    • 4 years ago
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    Yes, the coefficient of static friction is 0.31. Notice that these coefficients are unitless, because you have Newtons on both sides of the equation \( F = \mu N \) and force of friction \( F \) and the normal force \( F \) both have units of Newtons.

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