If an object is moving at $3 \frac{m}{s}$ $3 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{20}{g}$ $u_k=20 /g$ , how far will the object continue to move?

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If an object is moving at $3 \frac{m}{s}$ $3 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{20}{g}$ $u_k=20 /g$ , how far will the object continue to move?

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Answer 1 · 2016-05-10T09:13:24

$S = 0.225 m$ $S=0.225 m$

Explanation:

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$First; let's calculate the friction force:$ $"First; let's calculate the friction force:"$
$F_{f} = u_{k} \cdot G$ $F_f=u_k*G$
$G : weight of object G=m.g$ $G:"weight of object G=m.g"$
$u_{k} : coefficient of friction$ $u_k:" coefficient of friction"$

$F_{f} = u_{k} \cdot m \cdot g$ $F_f=u_k*m*g$

$F_{f} = \frac{20}{g} \cdot m \cdot g$ $F_f=20/g*m*g$

$F_{f} = 20 m$ $F_f=20m$

$a = \frac{F_{f}}{m}$ $a=F_f/m$

$a = \frac{20 m}{m} a = 20 \frac{m}{s^{2}} acceleration of object$ $a=(20m)/m " "a=20 m/s^2 " acceleration of object"$

$when object is stops ,the final velocity of its is zero$ $"when object is stops ,the final velocity of its is zero"$

$v_{f}^{2} = v_{i}^{2} - 2 \cdot a \cdot S$ $v_f^2=v_i^2-2*a*S$

$v_{f} : final velocity$ $v_f:"final velocity"$
$v_{i} : initial velocity$ $v_i:"initial velocity"$
$a : acceleration$ $a:"acceleration"$
$S: d i s p l a c e m e n t$ $"S:"displacement"$

$0 = 3^{2} - 2 \cdot 20 \cdot S$ $0=3^2-2*20*S$

$9 = 40 \cdot S$ $9=40*S$

$S = \frac{9}{40}$ $S=9/40$

$S = 0.225 m$ $S=0.225 m$

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If an object is moving at $3 \frac{m}{s}$ $3 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{20}{g}$ $u_k=20 /g$ , how far will the object continue to move?

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If an object is moving at $3 \frac{m}{s}$ $3 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{20}{g}$ $u_k=20 /g$ , how far will the object continue to move?