anonymous
  • anonymous
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 .
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
What is the magnitude of the resulting ac- celeration? Answer in units of m/s2
TuringTest
  • TuringTest
needs a picture...
anonymous
  • anonymous
ok

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anonymous
  • anonymous
|dw:1328461258569:dw|
anonymous
  • anonymous
|dw:1328461435693:dw|
anonymous
  • anonymous
then n * coefficient of friction which in this case is .152 would be f(k) = .874
anonymous
  • anonymous
|dw:1328461609676:dw|
anonymous
  • anonymous
i need to know if this is right!!!! please.
anonymous
  • anonymous
no one??????
TuringTest
  • TuringTest
I get 26.6m/s^2
TuringTest
  • TuringTest
\[\sum F_y=F\sin\theta-F_g=286\sin(27^{\circ})-20(9.8)\approx-57.3\]\[N\approx57.3\]\[\sum F_x=F+F\cos\theta-f\to a=\frac1m[F(1+\cos\theta)-\mu N]\]\[=\frac1{20}[286(1+\cos(27^{\circ}))-0.152(57.3)] \approx26.6m/s^2\]
anonymous
  • anonymous
where did the 27 degrees come from???
TuringTest
  • TuringTest
typo, meant 29 I can't see your answer so I couldn't check it
TuringTest
  • TuringTest
...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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