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

Apr 16, 2016

Force F required to compress a spring by x unit is
$F = k \cdot x$,where k = force constant.
Here $x = 0.4 m$
$k = 4 k g {s}^{-} 2$
hence Force exerted on the object of mass 1 kg is $F = 4 \frac{k g}{s} ^ 2 \times 0.4 m = 1.6 \frac{k g \cdot m}{s} ^ 2 = 1.6 N$
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
${\mu}_{s} = \frac{F}{N}$,where N is the normal reaction exerted on the body by the floor. Here $N = m g = 15 \cdot 9.8 N$
So
${\mu}_{s} = \frac{F}{N} = \frac{1.6}{9.8 \times 15} = 0.011$