Question A.2: A block of mass m slides over a frictionless surface in the ...

Using the work–energy theorem, we have

$\frac{1}{2} m v^{2}-\frac{1}{2} m v_{0}^{2}=-\mu_{k} m g \Delta x$

Add $1 \frac{1}{2} m v_{0}^{2}$ to both sides:

$\frac{1}{2} m v^{2}=\frac{1}{2} m v_{0}^{2}-\mu_{k} m g \Delta x$

Multiply both sides by 2/m:

$v^{2}=v_{0}^{2}-2 \mu_{k} g \Delta x$

Finally, take the square root of both sides. Because the block is sliding in the positive x-direction, the positive square root is selected.

$v=\sqrt{v_{0}^{2}-2 \mu_{k} g \Delta x}$