Question

A 65.0-kg woman steps off a 10.0-m diving platform and drops straight down into the water. 1) If she reaches a depth of 3.20 m, what is the average resistance force exerted on her by the water? Ignore any effects due to air resistance. (Express your answer to three significant figures in kN)

Answer 1

Answer:

F=2627.6N

Explanation:

The work done by this resistive force while traveling a distance d underwater would be:

$W=F.d=-Fd$

where the minus sign appears because the force is upwards and the displacement downwards.

This work is equal to the change of mechanical energy. At the diving plataform and underwater, when she stops moving, the woman has no kinetic energy, so all can be written in terms of her total change of gravitational potential energy:

$W=\Delta E=U_f-U_i=mgh_f-mgh_i=mg(h_f-h_i)$

Putting all together:

$F=-\frac{W}{d}=-\frac{mg(h_f-h_i)}{d}=-\frac{(65kg)(9.8m/s^2)(-3.2m-10m)}{3.2m}=2627.6N$

Answer 2

The average resistance force should be F=2627.6N

Calculation of the average resistance force:

Since the work done should be

W = f.d - fd

here. minus sign represent the force should be of upward direction and there is the fall of the displacement

Also

W = ΔE = U_f - U_i = mgh_f - mgh_i = mg(h_f - h_i)

Now the force should be

= -W/d

= -(65 kg) (9.8m/s^2) (-3.2m-10m)/ 3.2m

= 2627.6N

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