An object with a mass of $3 k g$ $3 kg$ is lying on a surface and is compressing a horizontal spring by $10 c m$ $10 cm$ . If the spring's constant is $8 \frac{k g}{s^{2}}$ $8 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

Question

1422 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-13T06:14:47

The coefficient of static friction is $= 0.027$ $=0.027$

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 $k = 8 k g s^{- 2}$ $k=8kgs^-2$

The compression is $x = 0.1 m$ $x=0.1m$

Therefore,

$F_{r} = 8 \cdot 0.1 = 0.8 N$ $F_r=8*0.1=0.8N$

The normal force is

$N = m g = 3 g N$ $N=mg=3gN$

The coefficient of static friction is

$μ_{s} = \frac{0.8}{3 g} = 0.027$ $mu_s=(0.8)/(3g)=0.027$

An object with a mass of 3kg 3 kg is lying on a surface and is compressing a horizontal spring by 10cm10 cm. If the spring's constant is 8kgs2 8 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?