An object with a mass of $15 k g$ $15 kg$ is lying on a surface and is compressing a horizontal spring by $40 c m$ $40 cm$ . If the spring's constant is $4 \frac{k g}{s^{2}}$ $4 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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An object with a mass of $15 k g$ $15 kg$ is lying on a surface and is compressing a horizontal spring by $40 c m$ $40 cm$ . If the spring's constant is $4 \frac{k g}{s^{2}}$ $4 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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Answer 1 · 2016-04-16T10:21:08

Force F required to compress a spring by x unit is
$F = k \cdot x$ $F=k*x$ ,where k = force constant.
Here $x = 0.4 m$ $x =0.4m$
$k = 4 k g s^{- 2}$ $k=4kgs^-2$
hence Force exerted on the object of mass 1 kg is $F = 4 \frac{k g}{s^{2}} \times 0.4 m = 1.6 \frac{k g \cdot m}{s^{2}} = 1.6 N$ $F=4(kg)/s^2xx0.4m=1.6(kg*m)/s^2=1.6N$
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 $N = m g = 15 \cdot 9.8 N$ $N =mg=15*9.8N$
So
$μ_{s} = \frac{F}{N} = \frac{1.6}{9.8 \times 15} = 0.011$ $mu_s=F/N=1.6/(9.8xx15)=0.011$

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An object with a mass of $15 k g$ $15 kg$ is lying on a surface and is compressing a horizontal spring by $40 c m$ $40 cm$ . If the spring's constant is $4 \frac{k g}{s^{2}}$ $4 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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An object with a mass of 15kg 15 kg is lying on a surface and is compressing a horizontal spring by 40cm40 cm. If the spring's constant is 4kgs2 4 (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 $15 k g$ $15 kg$ is lying on a surface and is compressing a horizontal spring by $40 c m$ $40 cm$ . If the spring's constant is $4 \frac{k g}{s^{2}}$ $4 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?