An object with a mass of $14 k g$ $14 kg$ is lying still on a surface and is compressing a horizontal spring by $1 m$ $1 m$ . If the spring's constant is $2 \frac{k g}{s^{2}}$ $2 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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Answer 1 · 2017-06-21T05:08:22

The coefficint of static friction is $= 0.015$ $=0.015$

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

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

Therefore,

$F_{r} = 2 \cdot 1 = 2 N$ $F_r=2*1=2N$

The normal force is

$N = m g = 14 g N$ $N=mg=14gN$

The coefficient of static friction is

$μ_{s} = \frac{2}{14 g} = 0.015$ $mu_s=(2)/(14g)=0.015$

An object with a mass of 14 kg14kg 14 kg is lying still on a surface and is compressing a horizontal spring by 1 m1m1 m. If the spring's constant is 2 (kg)/s^22kgs2 2 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?