An object with a mass of $5 k g$ $5 kg$ 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 $4 \frac{k g}{s^{2}}$ $4 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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Answer 1 · 2017-05-22T05:19:03

The coefficient of static friction is $= 0.12$ $=0.12$

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 $x = \frac{3}{2} m$ $x=3/2m$

Therefore,

$F_{r} = 4 \cdot \frac{3}{2} = 6 N$ $F_r=4*3/2=6N$

The normal force is

$N = m g = 5 g N$ $N=mg=5gN$

The coefficient of static friction is

$F_{r} = \frac{6}{5 g} = 0.12$ $F_r=(6)/(5g)=0.12$

An object with a mass of 5kg 5 kg is lying still on a surface and is compressing a horizontal spring by 32m3/2 m. If the spring's constant is 4kgs2 4 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?