The block has V-shaped grooves that interlock with the floor. As the block is pulled horizontally, it cannot simply slide flat. Instead, the “teeth” of the block must ride up the slopes of the floor’s grooves.

Let’s consider the motion over one full period of the groove, which corresponds to a horizontal distance $b$.

1. Rising Phase: The block is lifted vertically by height $h$ as it moves horizontally. This requires work to increase gravitational potential energy.
2. Falling Phase: After passing the peak, the block falls back down by height $h$.

We are given that collisions are “completely inelastic”. This implies that the kinetic energy gained by the block as it falls down the groove (converting potential energy back to kinetic) is dissipated upon impact at the bottom of the groove.

Therefore, for every horizontal distance $b$ traveled:

• Energy Input: Work is done by the pulling force $F_{\text{pull}}$.
  $W_{\text{in}} = F_{\text{pull}} \times b$.
• Energy Loss: The potential energy gained during the lift ($mgh$) is effectively discarded during the inelastic collision.
  $E_{\text{lost}} = mgh$.

Since the block moves at a constant average velocity, the work input must exactly equal the energy dissipated: $$ F_{\text{pull}} \cdot b = mgh $$ $$ F_{\text{pull}} = \frac{mgh}{b} $$

The effective coefficient of kinetic friction, $\mu_{\text{eff}}$, is defined by the relationship: $$ F_{\text{friction}} = \mu_{\text{eff}} N $$ Here, the pulling force $F_{\text{pull}}$ plays the role of the frictional resistance, and the normal force on the macroscopic horizontal surface is $N = mg$.

$$ \mu_{\text{eff}} = \frac{F_{\text{pull}}}{mg} $$ Substituting the expression for $F_{\text{pull}}$: $$ \mu_{\text{eff}} = \frac{\left( \frac{mgh}{b} \right)}{mg} $$ $$ \mu_{\text{eff}} = \frac{h}{b} $$

The suitable expression for the effective coefficient of kinetic friction is h/b.