Question

A 68.0-kg person jumps from rest off a 2.20-m-high tower straight down into the water. Neglect air resistance during the descent. She comes to rest 1.40 m under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is non-conservative.

Answer 1

Answer:

≅ 1714N

Explanation:

Start off by finding the total potential energy the person has at the top of the Tower.

PE = mass*g*Height

You can say the where the person stops is your reference level so at the 1.40m underwater there person has 0 potential energy. So the height is the Tower+1.40m

So for the person to stop 1.40m underwater they need to lose all of their energy to the water.

So work done by the water is the avg Force it exerts times the distance and the work done must equal the total enery the person has (which is the PE at the top of the tower). So:

Work = mgh

Work = F × d

F*d = m*g*h

m = 68kg

h = 2.20m + 1.40m = 3.60m

d = 1.40m

g = 9.8m/s²

$F = \frac{(68) \times(9.8)\times(3.60)}{1.40} \\\\F = 1713.6N$

≅ 1714N

Answer 2

Answer:

$F= 1333.767\,N$

Explanation:

The velocity of the swimmer just before touching the water is:

$v = -\sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.20\,m)}$

$v \approx 6.569\,\frac{m}{s}$

The average force exerted on the diver by the water is determined by the use of the Principle of Energy Conservation and the Work-Energy Theorem:

$(68\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (2.20\,m) +\frac{1}{2}\cdot (68\,kg)\cdot (6.569\,\frac{m}{s} )^{2}-F\cdot(2.20\,m) = 0\,J$

$F= 1333.767\,N$