A 6 gram coin is which is initially at rest is dropped from the observation deck of a skyscraper 300 meters above the street below.

Required:
a. What is the work done by gravity on the coin as it falls?
b. Because there is air resistance the coin was slowed and hit the ground with a final velocity of 60 m/s. What is the kinetic energy of the coin at this speed? Joules
c. How much work was lost to air resistance as the coin fell? Joules
d. What is the average force exerted on the coin due to the air resistance as it fell?

a) We must remind that situation with Earth-coin system must be represented by Principle of Energy Conservation and Work Energy Theorem. According to this latter, work done by gravity equals to the change in gravitational potential energy:

[tex]\Delta U_{g} = m\cdot g \cdot \Delta z[/tex] (Eq. 1)

Where:

[tex]\Delta U_{g}[/tex] - Change in gravitational potential energy, measured in joules.

[tex]m[/tex] - Mass, measured in kilograms.

[tex]g[/tex] - Gravitational accelerations, measured in meters per square second.

[tex]\Delta z[/tex] - Height of the skyscraper, measured in meters.

If we know that [tex]m = 0.006\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta z = 300\,m[/tex], the work done gravity on the coin is:

[tex]\Delta U_{g} = (0.006\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (300\,m)[/tex]

[tex]\Delta U_{g} = 17.653\,J[/tex]

The work done by gravity on the coin as it falls is 17.653 joules.

b) By definition of translation kinetic energy, we get the following model:

[tex]K = \frac{1}{2}\cdot m\cdot v^{2}[/tex]

Where:

[tex]K[/tex] - Kinetic energy of the coin right before hitting the street, measured in joules.

[tex]v[/tex] - Speed of the coin, measured in meters per second.

If we get that [tex]m = 0.006\,kg[/tex] and [tex]v = 60\,\frac{m}{s}[/tex], the kinetic energy at this speed is:

[tex]K = \frac{1}{2}\cdot (0.006\,kg)\cdot \left(60\,\frac{m}{s} \right)^{2}[/tex]

[tex]K = 10.8\,J[/tex]

The kinetic energy of the coin at a speed of 60 meters per second is 10.8 joules.

c) The work lost due to air resistance is obtained derived from Principle of Energy Conservation and Work-Energy Theorem:

[tex]W_{lost} = \Delta U_{g}-K[/tex] (Eq. 2)

Where [tex]W_{lost}[/tex] is the work lost due to air resistance, measured in joules.

If we know that [tex]\Delta U_{g} = 17.653\,J[/tex] and [tex]K = 10.8\,J[/tex], the work lost due to air resistance is:

[tex]W_{lost} = 17.653\,J-10.8\,J[/tex]

[tex]W_{lost} = 6.853\,J[/tex]

The work lost due to air resistance is 6.853 joules.

d) The average force exerted on the coin due to air resistance can be determined by applying definition of work, as air resistance force was antiparallel to the displacement of the coin. That is:

[tex]W_{lost} = F\cdot \Delta z[/tex]

[tex]F = \frac{W_{lost}}{\Delta z}[/tex] (Eq. 3)

Where [tex]F[/tex] is the average force exerted on the coin due to air resistance, measured in newtons.

If we know that [tex]W_{lost} = 6.853\,J[/tex] and [tex]\Delta z = 300\,m[/tex], then the average force exerted on the coin is:

[tex]F = \frac{6.853\,J}{300\,m}[/tex]

[tex]F = 0.023\,N[/tex]

The average force exerted on the coin due to air resistance as it fell is 0.023 newtons.