As shown in the figure, a block weighing 10 N travels down a smooth...

Given weight of the block, $W = mg = 10 \text{ N}$

Height of the track, $h = 1.0 \text{ m}$

Coefficient of friction, $\mu = 0.20$

By the work-energy theorem, the net work done on the block is equal to its change in kinetic energy.

Since the block starts from rest and finally comes to rest, the change in kinetic energy is zero, $\Delta K = 0$.

The work done by gravity as the block descends is:

$W_g = mgh = 10 \times 1.0 = 10 \text{ J}$

The work done by friction on the rough horizontal surface is:

$W_f = -\mu mg d = -0.20 \times 10 \times d = -2d$

Applying the work-energy theorem:

$W_g + W_f = \Delta K$

$10 - 2d = 0$

$2d = 10 \Rightarrow d = 5.0 \text{ m}$