How to calculate the work done by water on a diver?

Question

Background

Consider the a person with a mass of $m = 65 \text {kg}$ falling into water with a valocity of $14.007 \frac ms$.

At a depth of $3.1m$ the diver comes to a stop.

Question

How large is the force applied to the diver by the water? By this, we mean the sum of the force of friction and the buoyancy.

Thoughts

At the point of contact with the water, the diver has a kinetic energy of $E_k = \frac12 \cdot 65kg \cdot (14.007 m/s)^2 \approx 6.4 \cdot 10^3 J$.

The acceleration after entering the water is

$$a = \frac{v^2 - v_0^2}{2s} \approx -31.644 \frac m{s^2}$$

Let's call the work done by the water $R$. Then

$$\Sigma F = ma $$

$$ G-R = ma $$

$$R = m(g-a) = 65kg \cdot (9.81 m/s^2 + 31.644 m/s^2) \approx 2.69kN$$

which seems reasonable.

Alternative strategy doesn't work?

If I wanted instead to use $W = F\cdot s \Rightarrow F = \frac Ws$, how would I calculate the $W$-value to use here? I'm a bit shaky on the idea of work done by water here, when it's working against gravity. Is this equation useable here at all?

EDIT: Close votes

I see that someone has voted to close it as "belongs on another site on the network". I assume the target is Physics, but it was closed there because it's more on the calculation side, and not "general" enough.

Answer 1

From the equation $G-R = ma$, what you find cannot be a work. Indeed $R$ has the same units of measure of $G$ and $ma$, hence it's a force.

It's indeed the Archimede's force. If we took into account every force we would have

$$mg - F_A - S = ma$$

Where $F_A$ is the Archimede's force and $S$ is the Stokes' force. Here we do not deal with a spherical body hence the usual Stokes' law cannot be used. We refer in this case to the Drag Force, that is:

$$D = \frac12 \rho v^2 C_D\ A$$

Where $D$ is the drag force, which is by definition the force component in the direction of the flow velocity,

$\rho$ is the mass density of the fluid,

$v$ is the flow velocity relative to the object,

$A$ is the reference area, and

$C_D$ is the drag coefficient – a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag, in general $C_D$ depends on the Reynolds' number.

If we admit the Drag is negligible then we come up with

$$F_A = m(g-a)$$

$F_A$ is what you called $R$. Which again: is a force.

The work $W$ can be calculated via the innocent scalar product under some precise conditions, one of them is the force in play must be conservative. In the case of a sinking body, there is viscous friction (Stokes' / Drag Force indeed) which we neglected for the sake of simplicity.

For such a problem I would rather appeal to the Non-conservation of Energy, which states:

$$\mathcal{W} = \Delta E$$

Where $\mathcal{W}$ is the work done by the non-conservative forces in play (it would be the sum of the works of every NC force).

As usual

$$\Delta E = E_{\text{final}} - E_{\text{initial}} = (K + U)_{\text{final}} - (K + U)_{\text{initial}}$$

Where as usual $K = $ Kinetic Energy and $U = $ potential energy.

In your problem you have no idea of the initial height the person is falling from. But in the free falling, once at the groun (= sea surface here), all the potential Energy became Kinetic Energy. To analyse the second part of the problem, from the surface to when it stops under the sea we could use some analogous reasoning, even if I am not really sure it could properly work.