Answer:
[tex]K_f=1080J[/tex] and [tex]U_{gf}=539J[/tex].
Explanation:
From the conservation of the mechanical energy, we have that:
[tex]U_{g0}+K_0=U_{gf}+K_f[/tex]
Since the diver is in free fall, her initial velocity is zero. So her initial kinetic energy is also zero. Then, we get:
[tex]U_{g0}=U_{gf}+K_f[/tex]
We will first calculate the final kinetic energy. Solving for K_f, we have:
[tex]K_f=U_{g0}-U_{gf}=mgh_0-mgh_f=mg(h_0-h_f)\\\\K_f= (55.0kg)(9.81m/s^{2})(3.00m-1.00m)=1080J[/tex]
So the kinetic energy when the diver is 1.00m above the water is 1080J.
Finally, we solve for the final potential energy U_gf:
[tex]U_{gf}=U_{g0}-K_f=mgh_0-K_f\\\\U_{gf}=(55.0kg)(9.81m/s^{2})(3.00m)-1080J=539J[/tex]
In words, the gravitational potential energy when she is 1.00m above the water is 539J.
The gravitational potential energy of the diver is 539 J.
The kinetic energy of the diver is the diver is 1078 J.
The given parameters;
The final kinetic energy of the diver is the change in the potential energy of the diver;
[tex]K.E = \Delta P.E\\\\K.E = mg(h_1 - h_2)\\\\K.E = 55 \times 9.8 (3-1)\\\\K.E = 1078 \ J[/tex]
The gravitational potential energy of the diver is calculated as follows;
[tex]P.E_f = K.E - P.E_i\\\\P.E_f = 1078 \ - \ (55 \times 9.8 \times 1)\\\\P.E_f = 539 \ J[/tex]
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