An object with a mass of $16 k g$ $16 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{5}{3} m$ $5/3 m$ . If the spring's constant is $4 \frac{k g}{s^{2}}$ $4 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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An object with a mass of $16 k g$ $16 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{5}{3} m$ $5/3 m$ . If the spring's constant is $4 \frac{k g}{s^{2}}$ $4 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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Answer 1 · 2016-09-03T11:21:39

$μ = 0.0425$ $mu=0.0425$

Explanation:

Looking at the problem we can write the equilibrium force equation in horizontal direction:
$Friction force = Spring force$ $"Friction force = Spring force"$
So, $Friction Coefficient (μ) \cdot Reaction force (R) = Spring constant (k) \cdot Displacement (x)$ $"Friction Coefficient"(mu) * "Reaction force" (R) = "Spring constant" (k) * "Displacement" (x)$
$μ \cdot R = k \cdot x$ $mu * R = k*x$ ;or
$μ \cdot m g = k \cdot x$ $mu * mg=k*x$ ;or
$μ \cdot 16 \cdot 9.81 = 4 \cdot \frac{5}{3}$ $mu * 16*9.81=4*5/3$ ;or
$μ = 0.0425$ $mu=0.0425$
Thus the Friction coefficient is $μ = 0.0425$ $mu=0.0425$

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An object with a mass of $16 k g$ $16 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{5}{3} m$ $5/3 m$ . If the spring's constant is $4 \frac{k g}{s^{2}}$ $4 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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Explanation:

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An object with a mass of $16 k g$ $16 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{5}{3} m$ $5/3 m$ . If the spring's constant is $4 \frac{k g}{s^{2}}$ $4 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?