An object with a mass of $14 k g$ $14 kg$ 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 · 2018-01-18T12:56:32

The surface's coefficient of static friction, in this case, will be 0.0051, approximately.

Since the object is lying still on the surface, it is in equilibrium and, therefore, the net force in it must equal zero:

$F_{N} = 0$ $F_N = 0$ . (1)

Since it is placed by a horizontal spring, the friction force, $f$ $f$ , will also be in that direction, opposing to the spring's force.

Now, since the displacement $Deltax = 1m$ and $k = 7 N/m$ , then the elastic force is

$F = k . Deltax$ ;

$F = 7 . 1 = 7 N$ .

Now, from Equation (1), $f$ and $F$ must have equal intensities, which means that

$f = 7 N$ .

Now, since $f = mu . N$ , with $mu$ and $N$ being the surface's coefficient of static friction and $N$ the object's normal force, respectively, then:

$mu . N = 7$ .

Since in this case the normal force equals the object's weight, then:

$mu = 7/(9,81 . 14)$

$mu = 0.051$ , approximately (I have used #g = 9.81 m/s^2).

An object with a mass of 14 kg14kg 14 kg is lying still on a surface and is compressing a horizontal spring by 1 m1m1 m. If the spring's constant is 7 (kg)/s^27kgs2 7 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?