An object with a mass of $12 k g$ $12 kg$ is lying on a surface and is compressing a horizontal spring by $40 c m$ $40 cm$ . 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 · 2017-07-14T07:19:11

The coefficient of static friction is $= 0.014$ $=0.014$

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 $k = 4 k g s^{- 2}$ $k=4kgs^-2$

The compression is $x = 0.4 m$ $x=0.4m$

Therefore,

$F_{r} = 4 \cdot 0.4 = 1.6 N$ $F_r=4*0.4=1.6N$

The normal force is

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

The coefficient of static friction is

$μ_{s} = \frac{1.6}{12 g} = 0.014$ $mu_s=(1.6)/(12g)=0.014$

An object with a mass of 12kg 12 kg is lying on a surface and is compressing a horizontal spring by 40cm40 cm. If the spring's constant is 4kgs2 4 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?