Work & Coconuts

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

Assume +x and +y are positive.

Picture of the scenario: http://ift.tt/1xt0ONC

3. 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?

http://ift.tt/1xt0ONI

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