An object with a mass of $8 k g$ $8 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{5}{6} m$ $5/6 m$ . If the spring's constant is $9 \frac{k g}{s^{2}}$ $9 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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An object with a mass of $8 k g$ $8 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{5}{6} m$ $5/6 m$ . If the spring's constant is $9 \frac{k g}{s^{2}}$ $9 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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Answer 1 · 2017-08-28T06:40:35

0,0956

Explanation:

To don't move the assemblage, the friction force must be greater than spring one.
Force of the spring=
$F s = K X = 9 k \frac{g}{s^{2}} \times \frac{5}{6} m = 7, 5 N$ $Fs= KX=9 kg/s^2 xx 5/6 m = 7,5 N$
from friction force = $F a = f a \times m g$ $Fa= fa xx mg$ we have:
$f a = \frac{F a}{m g} = \frac{7, 5 N}{8 K g \times 9, 8 \frac{m}{s^{2}}} = 0, 0956$ $fa = (Fa)/(mg) = (7,5 N)/(8Kg xx 9,8 m/s^2) =0,0956$ that is the surface's minimum coefficient of static friction

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An object with a mass of $8 k g$ $8 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{5}{6} m$ $5/6 m$ . If the spring's constant is $9 \frac{k g}{s^{2}}$ $9 (kg)/s^2$ , 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 $8 k g$ $8 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{5}{6} m$ $5/6 m$ . If the spring's constant is $9 \frac{k g}{s^{2}}$ $9 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?