Question

A 55.0-kilogram diver falls freely from a diving platform that is 3.00 meters above the surface of the water in a pool. When she is 1.00 meter above the water, what are her gravitational potential energy and kinetic energy with respect to the water's surface?

Answer 1

Answer:

Ep = 539 and Ek = 1078 J

Explanation:

From the law of conservation of mechanical energy,

At any point in a mechanical system

Et = Ek +Ep................ Equation 1

Where Et = total Energy of the mechanical system, Ek = Kinetic energy of the system, Ep = Potential Energy of the system.

Et = mgH.................. Equation 2

Where m = mass of the diver, g = acceleration due to gravity, H = Height of the platform above the water surface

Given: m = 55 kg, H = 3 m, g = 9.8 m/s²

Substitute into equation 2

Et = 55(3)(9.8)

Et = 1617 J.

Also,

Ep = mgh............. Equation 3

Where h = height of the diver above the water.

Given: h = 1 m, m = 55 kg.

Substitute into equation 3

Ep = 55(1)(9.8)

Ep = 539 J.

make Ek the subject of equation 1

Ek = Et-Ep

Ek = 1617-539

Ek = 1078 J

Answer 2

Answer:

$K_f=1080J$ and $U_{gf}=539J$.

Explanation:

From the conservation of the mechanical energy, we have that:

$U_{g0}+K_0=U_{gf}+K_f$

Since the diver is in free fall, her initial velocity is zero. So her initial kinetic energy is also zero. Then, we get:

$U_{g0}=U_{gf}+K_f$

We will first calculate the final kinetic energy. Solving for K_f, we have:

$K_f=U_{g0}-U_{gf}=mgh_0-mgh_f=mg(h_0-h_f)\\\\K_f= (55.0kg)(9.81m/s^{2})(3.00m-1.00m)=1080J$

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:

$U_{gf}=U_{g0}-K_f=mgh_0-K_f\\\\U_{gf}=(55.0kg)(9.81m/s^{2})(3.00m)-1080J=539J$

In words, the gravitational potential energy when she is 1.00m above the water is 539J.