An object with a mass of $1 k g$ $1 kg$ is lying still on a surface and is compressing a horizontal spring by $1 m$ $1 m$ . If the spring's constant is $7 \frac{k g}{s^{2}}$ $7 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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An object with a mass of $1 k g$ $1 kg$ is lying still on a surface and is compressing a horizontal spring by $1 m$ $1 m$ . If the spring's constant is $7 \frac{k g}{s^{2}}$ $7 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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Answer 1 · 2016-03-06T09:16:55

Force F required to compress a spring by x unit is
$F = k \cdot x$ $F=k*x$ ,where k = force constant.
Here x =1m
$k =$ $k=$ 7kgs^-2
hence Force exerted on the object of mass 1 kg is $F = 7 \frac{k g}{s^{2}} \cdot 1 m = 7 \frac{k g \cdot m}{s^{2}} = 7 N$ $F=7(kg)/s^2*1m=7(kg*m)/s^2=7N$
This force is balanced by the then static frictional force as it is self adjusting one,

So the minimum value of the coefficient of static frictions
$μ_{s} = \frac{F}{N}$ $mu_s=F/N$ ,where N is the normal reaction exerted on the body by the floor. Here $N = m g = 1 \cdot 9.8 N = 9.8 N$ $N =mg=1*9.8N=9.8N$
So
$μ_{s} = \frac{F}{N} = \frac{7}{9.8} = 0.71$ $mu_s=F/N=7/9.8=0.71$

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An object with a mass of $1 k g$ $1 kg$ is lying still on a surface and is compressing a horizontal spring by $1 m$ $1 m$ . If the spring's constant is $7 \frac{k g}{s^{2}}$ $7 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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An object with a mass of 1kg 1 kg is lying still on a surface and is compressing a horizontal spring by 1m1 m. If the spring's constant is 7kgs2 7 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?

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An object with a mass of $1 k g$ $1 kg$ is lying still on a surface and is compressing a horizontal spring by $1 m$ $1 m$ . If the spring's constant is $7 \frac{k g}{s^{2}}$ $7 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?