An object with a mass of $7 k g$ $7 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{5}{4} m$ $5/4 m$ . 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 · 2017-04-15T09:20:05

The coefficient of static friction is $= 0.073$ $=0.073$

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The force of compression is

$F = k x = 4 \cdot \frac{5}{4} = 5 N$ $F=kx=4*5/4=5N$

The coefficient of kinetic friction is

$μ_{s} = \frac{F}{N}$ $mu_s=F/N$

The normal force is $N = m g = (7 g) N$ $N=mg=(7g)N$

So,

$μ_{s} = \frac{5}{7 g} = 0.073$ $mu_s=5/(7g)=0.073$

An object with a mass of 7 kg7kg7 kg is lying still on a surface and is compressing a horizontal spring by 5/4 m54m5/4 m. If the spring's constant is 4 (kg)/s^24kgs24 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?