An object with a mass of $1 k g$ $1 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{1}{3} m$ $1/3 m$ . If the spring's constant is $6 \frac{k g}{s^{2}}$ $6 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

Question

1489 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-10T14:03:20

The coefficient of static friction is $= 0.2$ $=0.2$

The mass is $m = 1 k g$ $m=1kg$

The compression of the spring is $x = \frac{1}{3} m$ $x=1/3m$

The spring constant is $k = 6 k g s^{- 2}$ $k=6kgs^-2$

The reaction of the spring is $R = k x = 6 \cdot \frac{1}{3} = 2 N$ $R=kx=6*1/3=2N$

The acceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

The normal reaction of the object is

$N = m g = 1 \cdot 9.8 = 9.8 N$ $N=mg=1*9.8=9.8N$

The coefficient of static friction is

$μ_{s} = \frac{R}{N} = \frac{2}{9.8} = 0.2$ $mu_s=R/N=2/9.8=0.2$

An object with a mass of 1kg1 kg is lying still on a surface and is compressing a horizontal spring by 13m1/3 m. If the spring's constant is 6kgs26 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?