If a $4 k g$ $4 kg$ object moving at $\frac{5}{4} \frac{m}{s}$ $5/4 m/s$ slows to a halt after moving $8 m$ $8 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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If a $4 k g$ $4 kg$ object moving at $\frac{5}{4} \frac{m}{s}$ $5/4 m/s$ slows to a halt after moving $8 m$ $8 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2016-06-26T05:11:38

$\approx 0.01$ $~~0.01$

Explanation:

Given
$m \to Mass of the object = 4 k g$ $m->"Mass of the object"=4kg$

$m \to Velocity of the object = \frac{5}{4} \frac{m}{s}$ $m->"Velocity of the object"=5/4m/s$
$g \to Acceleration due to gravity = 9.8 \frac{m}{s^{2}}$ $g->"Acceleration due to gravity"=9.8m/s^2$

$d \to Distance travelled before halt = 8 m$ $d->"Distance travelled before halt"=8m$

$μ_{k} \to Coefficient of kinetic friction = ?$ $mu_k->"Coefficient of kinetic friction"=?$

Considering law of conservation of energy we can say,
$work done against friction =KE of the object$ $"work done against friction =KE of the object"$

$\Rightarrow μ_{k} m g \times d = \frac{1}{2} \cdot m v^{2}$ $=>mu_kmgxxd=1/2*mv^2$

$\Rightarrow μ_{k} = \frac{v^{2}}{2 \cdot g \cdot d} = \frac{{(\frac{5}{4})}^{2}}{2 \cdot 9 .8 \cdot 8} \approx 0.01$ $=>mu_k=v^2/(2*g*d)=(5/4)^2/(2*9 .8*8)~~0.01$

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If a $4 k g$ $4 kg$ object moving at $\frac{5}{4} \frac{m}{s}$ $5/4 m/s$ slows to a halt after moving $8 m$ $8 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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If a $4 k g$ $4 kg$ object moving at $\frac{5}{4} \frac{m}{s}$ $5/4 m/s$ slows to a halt after moving $8 m$ $8 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?