An object with a mass of $18 k g$ $18 kg$ is lying on a surface and is compressing a horizontal spring by $30 c m$ $30 cm$ . If the spring's constant is $1 \frac{k g}{s^{2}}$ $1 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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Answer 1 · 2018-01-16T10:21:46

The coefficient of static friction is $= 0.0017$ $=0.0017$

The mass is $m = 18 k g$ $m=18kg$

The compression of the spring is $x = 0.3 m$ $x=0.3m$

The spring constant is $k = 1 k g s^{- 2}$ $k=1kgs^-2$

The reaction of the spring is $R = k x = 1 \cdot 0.3 = 0.3 N$ $R=kx=1*0.3=0.3N$

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

The normal reaction of the object is

$N = m g = 18 \cdot 9.8 = 176.4 N$ $N=mg=18*9.8=176.4N$

The coefficient of static friction is

$μ_{s} = \frac{R}{N} = \frac{0.3}{176.4} = 0.0017$ $mu_s=R/N=0.3/176.4=0.0017$

An object with a mass of 18 kg18kg 18 kg is lying on a surface and is compressing a horizontal spring by 30 cm30cm30 cm. If the spring's constant is 1 (kg)/s^21kgs2 1 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?