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A skier starts at the top of a frictionless slope and pushes off with a speed of 3.0 m/s. The elevation of the slope is 40 m. She skis down the slope to a valley with elevation 0.0 m and then glides to the peak of an adjacent slope that is at an elevation of 25 m. Calculate her speed at the second peak.

Sagot :

Given:

The speed of the skier at the top of the first peak is,

[tex]v_1=3.0\text{ m/s}[/tex]

The elevation of the slope is,

[tex]h_1=40\text{ m}[/tex]

She skis down the slope to a valley with an elevation of 0.0 m and then glides to the peak of an adjacent slope that is at an elevation of

[tex]h_2=25\text{ m}[/tex]

To find:

The speed at the second peak

Explanation:

The diagram can be depicted below as:

Using the conservation of the mechanical energy, we can write,

The total energy at peak 1 = The total energy at peak 2

So,

[tex]\begin{gathered} mgh_1+\frac{1}{2}mv_1^2=mgh_2+\frac{1}{2}mv_2^2 \\ v_2^2=2gh_1+v_1^2-2gh_2 \\ v_2=\sqrt{2gh_1+v_1^2-2gh_2} \end{gathered}[/tex]

Substituting the values we get,

[tex]\begin{gathered} v_2=\sqrt{2\times9.8\times40+(3.0)^2-2\times9.8\times25} \\ =\sqrt{303} \\ =17.4\text{ m/s} \end{gathered}[/tex]

Hence, the speed at the second peak is 17.4 m/s.

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