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anonymous

  • 4 years ago

The magnitude of each force is 286 N, the force on the right is applied at an angle 29 and the mass of the block is 20 kg. The coefficient of friction is 0.152. The acceleration of gravity is 9.8 m/s2 .

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  1. anonymous
    • 4 years ago
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    What is the magnitude of the resulting ac- celeration? Answer in units of m/s2

  2. TuringTest
    • 4 years ago
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    needs a picture...

  3. anonymous
    • 4 years ago
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    ok

  4. anonymous
    • 4 years ago
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    |dw:1328461258569:dw|

  5. anonymous
    • 4 years ago
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    |dw:1328461435693:dw|

  6. anonymous
    • 4 years ago
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    then n * coefficient of friction which in this case is .152 would be f(k) = .874

  7. anonymous
    • 4 years ago
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    |dw:1328461609676:dw|

  8. anonymous
    • 4 years ago
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    i need to know if this is right!!!! please.

  9. anonymous
    • 4 years ago
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    no one??????

  10. TuringTest
    • 4 years ago
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    I get 26.6m/s^2

  11. TuringTest
    • 4 years ago
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    \[\sum F_y=F\sin\theta-F_g=286\sin(27^{\circ})-20(9.8)\approx-57.3\]\[N\approx57.3\]\[\sum F_x=F+F\cos\theta-f\to a=\frac1m[F(1+\cos\theta)-\mu N]\]\[=\frac1{20}[286(1+\cos(27^{\circ}))-0.152(57.3)] \approx26.6m/s^2\]

  12. anonymous
    • 4 years ago
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    where did the 27 degrees come from???

  13. TuringTest
    • 4 years ago
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    typo, meant 29 I can't see your answer so I couldn't check it

  14. TuringTest
    • 4 years ago
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    ...I actually switched numbers partway through my calculation\[\sum F_y=F\sin\theta-F_g=286\sin(29^{\circ})-20(9.8)\approx-57.3\]\[N\approx57.3\]\[\sum F_x=F+F\cos\theta-f\to a=\frac1m[F(1+\cos\theta)-\mu N]\]\[=\frac1{20}[286(1+\cos(29^{\circ}))-0.152(57.3)] \approx26.37m/s^2\]

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