An object with a mass of $7 k g$ $7 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{5}{8} m$ $5 /8 m$ . If the spring's constant is $16 \frac{k g}{s^{2}}$ $16 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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Answer 1 · 2017-04-26T07:53:57

The coefficient of static friction is $= 0.15$ $=0.15$

The force of friction is

$F_r=k*Deltax$

The spring constant is $k=16kgs^-2$

The distance is $Deltax=5/8m$

Therefore,

$F_r=16*5/8=10N$

The coefficient of static friction is

$mu_s=F_r/N$

The normal force is $N=mg=(7g) N$

So,

$mu_s=(10)/(7g)=0.15$

An object with a mass of 7 kg7kg7 kg is lying still on a surface and is compressing a horizontal spring by 5 /8 m58m5 /8 m. If the spring's constant is 16 (kg)/s^216kgs216 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?