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}