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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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
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What is the magnitude of the resulting ac- celeration? Answer in units of m/s2
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then n * coefficient of friction which in this case is .152 would be f(k) = .874
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i need to know if this is right!!!! please.
no one??????
I get 26.6m/s^2
\[\sum F_y=F\sin\theta-F_g=286\sin(27^{\circ})-20(9.8)\approx-57.3\]\[N\approx57.3\]\[\sum F_x=F+F\cos\theta-f\to a=\frac1m[F(1+\cos\theta)-\mu N]\]\[=\frac1{20}[286(1+\cos(27^{\circ}))-0.152(57.3)] \approx26.6m/s^2\]
where did the 27 degrees come from???
typo, meant 29 I can't see your answer so I couldn't check it
...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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