An object with a mass of $18 k g$ $18 kg$ is lying on a surface and is compressing a horizontal spring by $60 c m$ $60 cm$ . If the spring's constant is $18 \frac{k g}{s^{2}}$ $18 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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Answer 1 · 2017-05-19T09:46:19

The coefficient of static friction is $= 0.061$ $=0.061$

The coefficient of static friction is

$μ_{s} = \frac{F_{r}}{N}$ $mu_s=F_r/N$

$F_{r} = k \cdot x$ $F_r=k*x$

The spring constant is $k = 18 k g s^{- 2}$ $k=18kgs^-2$

The compression is $x = 0.6 m$ $x=0.6m$

Therefore,

$F_{r} = 18 \cdot 0.6 = 10.8 N$ $F_r=18*0.6=10.8N$

The normal force is

$N = m g = 18 g N$ $N=mg=18gN$

The coefficient of static friction is

$F_{r} = \frac{10.8}{18 g} = 0.061$ $F_r=(10.8)/(18g)=0.061$

An object with a mass of 18kg 18 kg is lying on a surface and is compressing a horizontal spring by 60cm60 cm. If the spring's constant is 18kgs2 18 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?