A block is projected along a rough horizontal road at a speed of 10...

The initial velocity of the block is $u = 10 \text{ m/s}$.

The final velocity of the block is $v = 0 \text{ m/s}$.

The coefficient of kinetic friction is $\mu_k = 0.10$.

Taking the acceleration due to gravity as $g = 9.8 \text{ m/s}^2$, the deceleration of the block due to friction is:

$a = \mu_k g = 0.10 \times 9.8 = 0.98 \text{ m/s}^2$

Using the third equation of motion, $v^2 = u^2 - 2as$:

$0 = (10)^2 - 2(0.98)s$

$1.96s = 100$

$s = \dfrac{100}{1.96} = 51.02 \text{ m}$

Rounding to the nearest integer, the distance traveled before coming to a stop is $51 \text{ m}$.