Answer:
Minimum coefficient of kinetic friction between the surface and the block is [tex]\mu_k=\frac{kx}{2Mg}[/tex] .
Explanation:
Given:
Mass of the block = M
Spring constant = k
Distance pulled = x
According to the question:
We have to find the minimum co-efficient of kinetic friction between the surface and the block that will prevent the block from returning to its equilibrium with non-zero speed.
So,
From the FBD we can say that:
⇒ Normal force, [tex]N=Mg[/tex] ...equation(i)
⇒ Elastic potential energy, [tex]PE[/tex] = [tex]\frac{kx^2}{2}[/tex] ...equation (ii)
⇒ Frictional force, [tex]f[/tex] = [tex]\mu_kN[/tex] ...equation (iii)
⇒ Plugging (i) in (iii).
⇒ [tex]f=\mu_kMg[/tex]
Now,
⇒ As we know that the energy lost due to friction is equivalent to PE .
⇒ [tex]PE=fx[/tex] ...considering PE as [tex]mgh[/tex] or [tex]f(x)[/tex] .
Arranging the equation.
⇒ [tex]\frac{kx^2}{2}=\mu_k Mg (x)[/tex]
⇒ [tex]\frac{kx}{2}=\mu_k Mg[/tex] ...eliminating x from both sides.
⇒ [tex]\frac{kx}{2Mg}=\mu_k[/tex] ...dividing both sides wit Mg.
Minimum coefficient of kinetic friction between the surface and the block is [tex]\frac{kx}{2Mg}=\mu_k[/tex] .
