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3 years ago

6

Use this free body diagram to help you find the magnitude of the force F1 needed to keep this block in static equilibrium 15.3 N

ewton’s 36.3 Newton’s

1 answer:

bija089 [108]3 years ago

5 0

Do you have a picture of the diagram?

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Answer:

The coefficient of kinetic is

$u_{k}=0.59$

Explanation:

The forces in the axis 'x' and 'y' using law of Newton to find coefficient of kinetic friction

ΣF=m*a

ΣFy=W-N=0

ΣFy=Fn-Fu=m*a

$F_{u} =u_{k} *N\\F_{N}=25N\\N=W\\N=3.5kg*9.8\frac{m}{s^{2} }=34.3N$

$F_{N}-F_{u}=m*a\\F_{N}-u_{k}*N=m*a\\u_{k}*N=F_{N}-m*a\\u_{k}=\frac{F_{N}-m*a}{N}$

Now to find the coefficient can find the acceleration using equation of uniform motion accelerated

$v_{f} ^{2}=v_{o}^{2}+2*a(x_{f}-x_{o})\\x_{o}=0\\v_{o}=0\\v_{f} ^{2}=2*a*x_{f}\\a=\frac{v_{f} ^{2}}{2*a*x_{f}}\\ a=\frac{(1.53\frac{m}{s} )^{2}}{2*0.91m}\\a= 1.28 \frac{m}{s^{2} }$

So replacing the acceleration can fin the coefficient:

$u_{k}=\frac{F_{N}-m*a }{N}\\u_{k}=\frac{25N-(3.5kg*1.28\frac{m}{s^{2}} }{34.3N} \\u_{k}=0.59$

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