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

Answer 1

Elena Albarran

The coefficint of static friction is #=0.015#

The coefficient of static friction is

#mu_s=F_r/N#

#F_r=k*x#

The spring constant is #k=2kgs^-2#

The compression is #x=1m#

Therefore,

#F_r=2*1=2N#

The normal force is

#N=mg=14gN#

The coefficient of static friction is

#mu_s=(2)/(14g)=0.015#

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Answer 2

Benjamin Asiedu

To find the minimum value of the surface's coefficient of static friction, we use the formula:

( \mu_s = \frac{kx}{mg} )

where:

( \mu_s ) is the coefficient of static friction
( k ) is the spring constant (2 (kg)/s^2)
( x ) is the compression of the spring (1 m)
( m ) is the mass of the object (14 kg)
( g ) is the acceleration due to gravity (9.8 m/s^2)

Plugging in the values:

( \mu_s = \frac{2 \times 1}{14 \times 9.8} )

( \mu_s = \frac{2}{137.2} )

( \mu_s \approx 0.015 )

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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.