Answer:
Approximately [tex]5.05\; \rm m \cdot s^{-1}[/tex].
Explanation:
The sled gains gravitational potential energy (GPE) as it moves up the incline. Since energy conserves, that energy has to come from the initial kinetic energy (KE). Additionally, since the incline is frictionless, all of the initial KE would be converted to GPE. That gives the equation
[tex]\displaystyle \frac{1}{2} \, m \cdot {v_\text{initial}}^{2} = \text{Initial KE} = \text{Final GPE} = m \cdot g \cdot h[/tex],
where
Although the mass [tex]m[/tex] of the sled isn't given, it appears on both sides of this equation. Divide both sides by [tex]m[/tex] to eliminate it.
[tex]\displaystyle \frac{1}{2} \, v^{2} = g \cdot h[/tex].
Rearrange the equation to isolate [tex]v[/tex]:
[tex]v^2 = 2 \, g \cdot h[/tex],
[tex]v = \sqrt{2 \, g \cdot h[/tex].
Evaluate the expression to obtain the value of [tex]v[/tex]:
[tex]v = \sqrt{2\times 9.81 \times 1.3} \approx 5.05\; \rm m \cdot s^{-1}[/tex].
