An object with a mass of $12 k g$ $12 kg$ is lying still on a surface and is compressing a horizontal spring by $3 m$ $3 m$ . 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 $12 k g$ $12 kg$ is lying still on a surface and is compressing a horizontal spring by $3 m$ $3 m$ . 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 · 2016-03-19T09:19:21

$μ = \frac{36}{12 \times 10} = 0.3$ $mu =36/(12xx10)=0.3$

Explanation:

We know that the gravitational Force applied or normal reaction $F_{N} = m \cdot g = 12 \cdot 10 N$ $F_N= m*g=12*10 N$ [ Taking g= 10 m/s^2]
force of limiting value of static friction F= $μ \cdot F_{N} = μ \cdot 12 \cdot 10 N$ $mu* F_N=mu*12*10N$
Again $F = \frac{1}{2} k x^{2} = \frac{1}{2} \cdot 4 \cdot 3^{2} = 36 N$ $F =1/2kx^2= 1/2*4*3^2=36N$ , where k = force constant = $4 \frac{k g}{s^{2}}$ $4(kg)/s^2$ x = amount of compression of spring '
So
$μ \cdot 12 \cdot 10 = 36$ $mu*12*10=36$
$μ = \frac{36}{12 \times 10} = 0.3$ $mu =36/(12xx10)=0.3$

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An object with a mass of $12 k g$ $12 kg$ is lying still on a surface and is compressing a horizontal spring by $3 m$ $3 m$ . 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 12kg12 kg is lying still on a surface and is compressing a horizontal spring by 3m3 m. If the spring's constant is 4kgs24 (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 $12 k g$ $12 kg$ is lying still on a surface and is compressing a horizontal spring by $3 m$ $3 m$ . 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?