Question

An object with a mass of `12 kg` is lying still on a surface and is compressing a horizontal spring by `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

`mu =36/(12xx10)=0.3`

Explanation:

We know that the gravitational Force applied or normal reaction `F_N= m*g=12*10 N` [ Taking g= 10 m/s^2]
force of limiting value of static friction F= `mu* F_N=mu*12*10N`
Again `F =1/2kx^2= 1/2*4*3^2=36N`, where k = force constant = `4(kg)/s^2` x = amount of compression of spring '
So
`mu*12*10=36`
`mu =36/(12xx10)=0.3`