Solving a Coconut's Fall: Speed, Work, & Air Resistance

[STEMucator]

Homework Statement

A coconut, with $m = 2.30 kg$, falls from rest from a tree which is $h = 35.0 m$ high. Take $g = 9.78 m/s^2$.

1. What speed does the coconut hit the ground? Ignore air resistance.

2. If it actually hits the ground with a speed of $23.9 m/s$, how much work was done by air resistance?

3. Air resistance actually varies with speed, but it is possible to calculate an average value for the air resistance using (1) and (2). Find this average force.

Homework Equations

Assume +x and +y are positive.

Picture of the scenario: http://gyazo.com/80cbb4b9e840106dc893e99eb1b76020

The Attempt at a Solution

Place a datum plane at the bottom with which to reference potential energy.

1. I simply used conservation. At the top, there is only potential. At the bottom, all the energy is kinetic, hence:

$U_{g_i} = K_f$
$mgh_i = \frac{1}{2} mv_f^2$
$(2.30)(9.78)(35.0) = \frac{1}{2} (2.30) v_f^2$
$v_f = \sqrt{2(9.78)(35.0)} = 26.165 m/s = 26.2 m/s$​

2. I was thinking I should use the work-kinetic energy theorem for this.

$W = \Delta K = K_f - K_i = \frac{1}{2} (2.30) (23.9)^2 = 656.892 J = 657 J$​

3. From part (1), we know that the work done by the air resistance is:

$W_1 = \Delta K = K_f - K_i = \frac{1}{2} (2.30) (26.165)^2 = 787.298 J = 787 J$​

From part (2), we know $W_2 = 656.892 J = 657 J$ hence the average work done by air resistance can be found:

$W_{avg} = \frac{W_1 + W_2}{2} = 722.095 J = 722 J$​

Hence we can find the average force of the air resistance (assuming it's not variable):

$W_{avg} = F_{air} d cos(\theta)$
$722.095 = F_{air} (- 35.0) cos(180°)$
$F_{air} = \frac{722.095}{35.0} = 20.631 N = 20.6 N$​

The positive answer indicates that the air resistance acts in the +y direction.

Does this look okay?

 

[Doc Al]

2. I was thinking I should use the work-kinetic energy theorem for this.

$W = \Delta K = K_f - K_i = \frac{1}{2} (2.30) (23.9)^2 = 656.892 J = 657 J$​

This is the net work done when air resistance acts.

3. From part (1), we know that the work done by the air resistance is:

$W_1 = \Delta K = K_f - K_i = \frac{1}{2} (2.30) (26.165)^2 = 787.298 J = 787 J$​

This is the net work done when air resistance does not act.

Revise your answers accordingly.

 

[STEMucator]

This is the net work done when air resistance acts.


This is the net work done when air resistance does not act.

Revise your answers accordingly.

Thank you, I have face-palmed accordingly.

So when air resistance does not act, the work done is $W_1 = 787 J$.

When air resistance does act, the work done is $W_2 = 657 J$.

The difference between the work done when there is no air resistance and when there is air resistance is $W_1 - W_2 = 130 J$.

Finding the average force of the air resistance now yields $F_{air} = 3.714 N = 3.71 N$.

 

[Doc Al]

Now you're cooking. :thumbs:

 

[HalfLostAndSearching]

I would say that your solution looks mostly correct. However, there are a few things that could be improved upon.

Firstly, in part 1, you have correctly used the conservation of energy principle to find the final velocity of the coconut. However, it would be more accurate to use the formula for gravitational potential energy, $U_g = mgh$, instead of $U_{g_i}$, which is not a commonly used notation.

In part 2, you have correctly used the work-kinetic energy theorem to find the work done by air resistance. However, it would be more accurate to use the formula $W = \int F_{air} dx$, where $F_{air}$ is the variable force of air resistance and $dx$ is the displacement of the coconut. This takes into account the fact that air resistance is not a constant force and varies with the speed of the coconut.

In part 3, you have correctly calculated the average work done by air resistance and used it to find the average force of air resistance. However, it would be more accurate to use the formula $F_{avg} = \frac{\Delta p}{\Delta t}$, where $\Delta p$ is the change in momentum of the coconut and $\Delta t$ is the time it takes for the coconut to fall. This takes into account the fact that air resistance is a force that acts over a period of time, not just at the moment of impact.

Overall, your solution is a good start, but with a few improvements, it could be even more accurate and scientifically rigorous.