An object with a mass of $7 k g$ $7 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{5}{4} m$ $5/4 m$ . 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 · 2017-12-17T12:43:56

The coefficient of static friction is $= 0.055$ $=0.055$

The mass is $m = 7 k g$ $m=7kg$

The compression of the spring is $x = \frac{5}{4} m$ $x=5/4m$

The spring constant is $k = 3 k g s^{- 2}$ $k=3kgs^-2$

The reaction of the spring is $R = k x = 3 \cdot \frac{5}{4} = \frac{15}{4} = 3.75 N$ $R=kx=3*5/4=15/4=3.75N$

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

The normal reaction of the object is $N = m g = 7 \cdot 9.8 = 68.6 N$ $N=mg=7*9.8=68.6N$

The coefficient of static friction is

$μ_{s} = \frac{R}{N} = \frac{3.75}{68.6} = 0.055$ $mu_s=R/N=3.75/68.6=0.055$

An object with a mass of 7kg7 kg is lying still on a surface and is compressing a horizontal spring by 54m5/4 m. If the spring's constant is 3kgs23 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?