An object with a mass of $4 k g$ $4 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{1}{4} m$ $1/4 m$ . If the spring's constant is $3 \frac{k g}{s^{2}}$ $3 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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Answer 1 · 2017-06-03T11:46:13

The coefficient of static friction is $= 0.02$ $=0.02$

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 $k = 3 k g s^{- 2}$ $k=3kgs^-2$

The compression is $x = \frac{1}{4} m$ $x=1/4m$

Therefore,

$F_{r} = 3 \cdot \frac{1}{4} = \frac{3}{4} N$ $F_r=3*1/4=3/4N$

The normal force is

$N = m g = 4 g N$ $N=mg=4gN$

The coefficient of static friction is

$F_{r} = \frac{\frac{3}{4}}{4 g} = 0.02$ $F_r=(3/4)/(4g)=0.02$

An object with a mass of 4kg4 kg is lying still on a surface and is compressing a horizontal spring by 14m1/4 m. If the spring's constant is 3kgs23 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?