An object with a mass of $16 k g$ $16 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{7}{8} m$ $7/8 m$ . If the spring's constant is $15 \frac{k g}{s^{2}}$ $15 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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Answer 1 · 2017-05-19T08:28:09

The coeficient of friction is $= 0.084$ $=0.084$

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 $k = 15 k g s^{- 2}$ $k=15kgs^-2$

The compression is $x = \frac{7}{8} m$ $x=7/8m$

Therefore,

$F_{r} = 15 \cdot \frac{7}{8} = \frac{105}{8} N$ $F_r=15*7/8=105/8N$

The normal force is

$N = m g = 16 g N$ $N=mg=16gN$

The coefficient of static friction is

$F_{r} = \frac{\frac{105}{8}}{16 g} = \frac{13.125}{16 g} = 0.084$ $F_r=(105/8)/(16g)=13.125/(16g)=0.084$

An object with a mass of 16 kg16kg16 kg is lying still on a surface and is compressing a horizontal spring by 7/8 m78m7/8 m. If the spring's constant is 15 (kg)/s^215kgs215 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?