The specialty of an athlete on the women's track team is the pole vault. She has a mass of 60.0 kg and her approach speed is 9.20 m/s. When she is directly above the bar, her speed is 2.40 m/s. Neglecting air resistance and any energy absorbed by the pole, determine the amount she has raised herself as she crosses the bar.

Respuesta :

Answer:

She raised approximately 4 meters as she crosses the bar

Explanation:

Energy Conservation

In the absence of air resistance and any other energy-absorbing events, the total mechanical energy is conserved. Recall the mechanical energy is the sum of the gravitational potential energy and the kinetic energy:

[tex]\displaystyle M=m.g.h+\frac{1}{2}m.v^2[/tex]

When our athlete is at the ground level her height is 0, thus she has only kinetic energy given by the speed she's at when starting the jump:

[tex]\displaystyle M_1=m.g.(0)+\frac{1}{2}m.v^2=\frac{1}{2}m.v^2[/tex]

Later when she's directly above the bar, she has both energies since some speed remains at a certain height h we must calculate. The mechanical energy is

[tex]\displaystyle M_2=m.g.h+\frac{1}{2}m.v'^2[/tex]

Equating both energies

[tex]\displaystyle m.g.h+\frac{1}{2}m.v'^2=\frac{1}{2}m.v^2[/tex]

Simplifying by m and rearranging

[tex]\displaystyle m.g.h=\frac{1}{2}m.v^2-\frac{1}{2}m.v'^2=\frac{1}{2}(v^2-v'^2)[/tex]

Solving for h

[tex]\displaystyle h=\frac{(v^2-v'^2)}{2g}=\frac{(9.2^2-2.4^2)}{2\cdot 9.8}[/tex]

[tex]\boxed{h=4.02\ m}[/tex]

She raised approximately 4 meters as she crosses the bar

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