Question

A 61.0-kg person jumps from rest off a 10.0-m-high tower straight down into the water. Neglect air resistance. She comes to rest 3.00 m under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is nonconservative.

Answer 1

Answer:

Explanation:

In this case, law of conservation of energy will be implemented. It states that "the energy of the system remains conserved until or unless some external force act on it. Energy of the system may went through the conversion process like kinetic energy into potential and potential into kinetic energy.But their total always remain the same in conserved systems."

Given data:

Height of tower = 10.0 m

Depth of the pool = 3.00 cm

Mass of person = 61.0 kg

Solution:

Initial energy = Final energy

$U_{i} = (K.E) + U_{f}$

As the person was at height initially so it has the potential energy only.

$mg(h_{1} +h_{2}) = K.E + mgh_{2}$

$K.E = mgh_{1}$

$K.E = (61.0)(9.8)(10)\\K.E = 5978 J$

Lets find out the magnitude of the force that the water is exerting on the diver.

W =ΔK.E

$F.h_{2} = 5978$

$F = \frac{5978}{3}$

F = 1992.67 N