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anonymous

  • 4 years ago

a 5kg block is pushed along a level surface by a 40N force at an angle of 30 degrees below the surface. The force of friction action on the block is 10N. What is the acceleration of the block? Normal Force? and Coefficient of kinetic friction.

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  1. NotTim
    • 4 years ago
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    As always, I suggest drawing out a diagram, and then creating a variable list.

  2. anonymous
    • 4 years ago
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    well the triangle should be on the other side...hol d on

  3. anonymous
    • 4 years ago
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    |dw:1328577203656:dw|

  4. NotTim
    • 4 years ago
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    Acceleration- F=ma (Rearrange the formula; F uses an angle, and also has a "resistant" frictional force, so remember SOHCAHTOA)

  5. NotTim
    • 4 years ago
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    I think. Don't take my word for it.

  6. anonymous
    • 4 years ago
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    n=mg-Fsintheta

  7. NotTim
    • 4 years ago
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    Whoops. Gotta go...Maybe someone else may help. There's 2 other people here.

  8. NotTim
    • 4 years ago
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    That looks right, but what's n?

  9. NotTim
    • 4 years ago
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    I'll be back in 30min.

  10. anonymous
    • 4 years ago
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    normal force. i think you have to calculate normal fornce to find acceleration

  11. anonymous
    • 4 years ago
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    ok ill still be here. :/

  12. anonymous
    • 4 years ago
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    \[(F)\cos(30)= Fx\] \[f=\mu F _{N}\] |dw:1328578095405:dw|

  13. anonymous
    • 4 years ago
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    so what is -f in this drawing....

  14. anonymous
    • 4 years ago
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    is it all the forces combined? kinetic friction?

  15. anonymous
    • 4 years ago
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    -f is force of friction that is being Fx is component of the force applied by you so net force in horizontal direction would be |dw:1328602865107:dw| solve for than a gives accel;eration

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