An object with a mass of $12 k g$ $12 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{3}{2} m$ $3/2 m$ . 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 · 2017-04-27T05:44:26

The coefficient of static friction is $= 0.026$ $=0.026$

The force of friction is

$F_r=k*Deltax$

The spring constant is $k=2kgs^-2$

The distance is $Deltax=3/2m$

Therefore,

$F_r=2*3/2=3N$

The coefficient of static friction is

$mu_s=F_r/N$

The normal force is $N=mg=(12g) N$

So,

$mu_s=(3)/(12g)=0.026$

An object with a mass of 12 kg12kg 12 kg is lying still on a surface and is compressing a horizontal spring by 3/2 m32m3/2 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?