Question

A block of mass M on a horizontal surface is connected to the end of a massless spring of spring constant k. The block is pulled a distance x from equilibrium and when released from rest, the block moves toward equilibrium. What minimum coefficient of kinetic friction between the surface and the block would prevent the block from returning to equilibrium with non-zero speed

Answer 1

Answer:

Minimum coefficient of kinetic friction between the surface and the block is $\mu_k=\frac{kx}{2Mg}$ .

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, $N=Mg$                                   ...equation(i)

⇒ Elastic potential energy, $PE$ = $\frac{kx^2}{2}$                ...equation (ii)

⇒ Frictional force, $f$ = $\mu_kN$                                 ...equation (iii)

⇒ Plugging (i) in (iii).

⇒ $f=\mu_kMg$

Now,

⇒ As we know that the energy lost due to friction is equivalent to PE .

⇒ $PE=fx$                     ...considering PE as $mgh$ or $f(x)$ .

   Arranging the equation.

⇒ $\frac{kx^2}{2}=\mu_k Mg (x)$

⇒ $\frac{kx}{2}=\mu_k Mg$                 ...eliminating x from both sides.

⇒ $\frac{kx}{2Mg}=\mu_k$                    ...dividing both sides wit Mg.

Minimum coefficient of kinetic friction between the surface and the block is $\frac{kx}{2Mg}=\mu_k$ .