An object with a mass of $4 k g$ $4 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{1}{4} m$ $1/4 m$ . If the spring's constant is $6 \frac{k g}{s^{2}}$ $6 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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An object with a mass of $4 k g$ $4 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{1}{4} m$ $1/4 m$ . If the spring's constant is $6 \frac{k g}{s^{2}}$ $6 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

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Answer 1 · 2017-01-07T00:17:48

$μ_{s} \approx 0.04$ $mu_s~~0.04$

Explanation:

We will solve this first by setting up a force diagram and writing out sum of forces statements.

Where $\to n$ $vecn$ is the normal force, ${\to F}_{g}$ $vecF_g$ is the force of gravity, ${\to F}_{s p}$ $vecF_(sp)$ is the force of the spring, and ${\to f}_{s}$ $vecf_s$ is the force of static friction. I will define rightward as positive.

Statements of the net force:

$\sum F_{x} = {\to F}_{s p} - {\to f}_{s} = m {\to a}_{x}$ $sumF_x=vecF_(sp)-vecf_s=mveca_x$

$\sum F_{y} = \to n - {\to F}_{g} = m {\to a}_{y}$ $sumF_y=vecn-vecF_g=mveca_y$

The object is not moving, and experiences no net acceleration in either $x$ $x$ - or $y$ $y$ -direction.

$sumF_x=vecF_(sp)-vecf_s=cancel(mveca_x)=vec0$

$sumF_y=vecn-vecF_g=cancel(mveca_y)=vec0$

We can rearrange these statements to find:

$vecF_(sp)=vecf_s$

$vecn=vecF_g$

Where $vecF_g=mg$

The equation for the frictional force is $vecf=muvecn$ . Therefore, we have:

$vecF_(sp)=mumg$

The equation for the spring force is given by Hooke's law as $vecF_(sp)=k(Deltas)$ , where $k$ is the spring constant and $Deltas$ is the displacement from equilibrium (how far the spring is stretched or compressed).

$k(Deltas)=mumg$

Solving for $mu$ , the coefficient of friction:

$mu=(k(Deltas))/(mg)$

Using our known values, we can calculuate $mu$ :

$mu=((6N/m)(1/4m))/(4kg*9.8m/s^2)$

$mu=0.038...~~0.04$

Therefore, the minimum coefficient of static friction to keep the object still is $~~0.04$ .

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An object with a mass of $4 k g$ $4 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{1}{4} m$ $1/4 m$ . If the spring's constant is $6 \frac{k g}{s^{2}}$ $6 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?

1 Answer

Explanation:

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1 Answer

Explanation:

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An object with a mass of $4 k g$ $4 kg$ is lying still on a surface and is compressing a horizontal spring by $\frac{1}{4} m$ $1/4 m$ . If the spring's constant is $6 \frac{k g}{s^{2}}$ $6 (kg)/s^2$ , what is the minimum value of the surface's coefficient of static friction?