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?

Answer 1

Force F required to compress a spring by x unit is #F=k*x#,where k = force constant. Here #x =0.4m# #k=4kgs^-2# hence Force exerted on the object of mass 1 kg is #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 #mu_s=F/N#,where N is the normal reaction exerted on the body by the floor. Here #N =mg=15*9.8N# So #mu_s=F/N=1.6/(9.8xx15)=0.011#

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Answer 2

Sebastian Acosta

The minimum value of the surface's coefficient of static friction is 0.5.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.