An object with a mass of $12 k g$ $12 kg$ is lying still on a surface and is compressing a horizontal spring by $1 m$ $1 m$ . If the spring's constant is $15 \frac{k g}{s^{2}}$ $15 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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Answer 1 · 2017-07-12T06:22:57

The coefficient of static friction is $= 0.13$ $=0.13$

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 $k = 15 k g s^{- 2}$ $k=15kgs^-2$

The compression is $x = 1 m$ $x=1m$

Therefore,

$F_{r} = 15 \cdot 1 = 15 N$ $F_r=15*1=15N$

The normal force is

$N = m g = 12 g N$ $N=mg=12gN$

The coefficient of static friction is

$μ_{s} = \frac{15}{12 g} = 0.13$ $mu_s=(15)/(12g)=0.13$

An object with a mass of 12kg 12 kg is lying still on a surface and is compressing a horizontal spring by 1m1 m. If the spring's constant is 15kgs2 15 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?