An object with a mass of #12 kg# is lying still on a surface and is compressing a horizontal spring by #1/3 m#. 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 =1/3m# #k=4kgs^-2# hence Force exerted on the object of mass 1 kg is #F=4(kg)/s^2xx1/3m=4/3(kg*m)/s^2=4/3N# 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=12*9.8N# So #mu_s=F/N=(4/3)/(9.8xx12)=0.011#
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Answer 2

The minimum value of the surface's coefficient of static friction is ( \mu_s = \frac{kx}{mg} ), where ( k ) is the spring constant, ( x ) is the compression of the spring, ( m ) is the mass of the object, and ( g ) is the acceleration due to gravity.

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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.

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.

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.

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.

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