An object with a mass of $12 k g$ $12 kg$ is lying on a surface and is compressing a horizontal spring by $50 c m$ $50 cm$ . If the spring's constant is $3 \frac{k g}{s^{2}}$ $3 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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Answer 1 · 2017-06-13T10:05:21

The coefficient of static friction is $= 0.013$ $=0.013$

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 $k = 3 k g s^{- 2}$ $k=3kgs^-2$

The compression is $x = 0.5 m$ $x=0.5m$

Therefore,

$F_{r} = 3 \cdot 0.5 = 1.5 N$ $F_r=3*0.5=1.5N$

The normal force is

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

The coefficient of static friction is

$μ_{s} = \frac{1.5}{12 g} = 0.013$ $mu_s=(1.5)/(12g)=0.013$

An object with a mass of 12kg 12 kg is lying on a surface and is compressing a horizontal spring by 50cm50 cm. If the spring's constant is 3kgs2 3 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?