The only force acting on the block is the force of kinetic friction. This force is given by:

$$F_k=\mu_kmg$$

where \(\mu_k\) is the coefficient of kinetic friction, \(\(mg\) is the weight of the block.

Step 2: Write down Newton's second law for the block

In the horizontal direction, Newton's second law for the block is given by:

$$ma=-\mu_k mg$$

Where \(a\) is the acceleration of the block in the \(x\) direction.

Step 3: Solve the equation of motion for the block

We can solve the equation of motion for the block by using the following formula:

$$v_f^2=v_i^2+2ad$$

where \(v_f\) is the final speed of the block, \(v_i\) is the initial speed of the block, \(a\) is the acceleration of the block, and \(d\) is the distance traveled by the block.

In this case, the final speed of the block is 0 m/s, the initial speed of the block is \(v\), the acceleration of the block is \(-\mu_k g\), and the distance traveled by the block is \(d\).

Substituting these values into the formula, we get:

$$0^2=v^2+2(-\mu_k g)d$$

Solving for \(d\), we get:

$$d=\frac{v^2}{2\mu_k g}$$

Therefore, the block will travel a distance of \(\frac{v^2}{2\mu_k g}\) across the horizontal surface before coming to a stop.