An object with a mass of $5 k g$ $5 kg$ is lying still on a surface and is compressing a horizontal spring by $2 m$ $2 m$ . 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?

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An object with a mass of $5 k g$ $5 kg$ is lying still on a surface and is compressing a horizontal spring by $2 m$ $2 m$ . 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?

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Answer 1 · 2016-04-08T12:52:00

Force F required to compress a spring by x unit is
$F = k \cdot x$ $F=k*x$ ,where k = force constant.
Here x =2m
$k = 8 k g s^{- 2}$ $k=8kgs^-2$
hence Force exerted on the object of mass 1 kg is $F = 8 \frac{k g}{s^{2}} \cdot 2 m = 16 \frac{k g \cdot m}{s^{2}} = 16 N$ $F=8(kg)/s^2*2m=16(kg*m)/s^2=16N$
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 $N = m g = 5 \cdot 9.8 N$ $N =mg=5*9.8N$
So
$μ_{s} = \frac{F}{N} = \frac{16}{9.8 \times 5} = 0 . .33$ $mu_s=F/N=16/(9.8xx5)=0..33$

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An object with a mass of $5 k g$ $5 kg$ is lying still on a surface and is compressing a horizontal spring by $2 m$ $2 m$ . 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?

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An object with a mass of 5kg 5 kg is lying still on a surface and is compressing a horizontal spring by 2m2 m. If the spring's constant is 8kgs2 8 (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 $5 k g$ $5 kg$ is lying still on a surface and is compressing a horizontal spring by $2 m$ $2 m$ . 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?