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: