Question

The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion.

Assume that the North American continent can be represented by a slab of rock 4450 km on a side and 31 km deep and that the rock has an average mass density of 2620 kg/m3.
The continent is moving at the rate of about 1 cm/year.
What is the mass of the continent? Answer in units of kg.
(part 2 of 3)
What is the kinetic energy of the continent? Answer in units of J.
(part 3 of 3)
A jogger (of mass 77 kg) has the same kinetic energy as that of the continent.
What would his speed be? Answer in units of m/s.

Answer 1

1) The mass of the continent is approximately
`1.608* 10^(21)` kilograms.

2) The kinetic energy of the continent is approximately
`8.04* 10^(16)` joules.

3) The speed of the 77 kg-jogger would be approximately
`45.698* 10^(6)` meters per second.

Step-by-step explanation:

1) The mass of the North American continent can be estimated by using the following formula under the assumption that rock has an uniform density:

`m = \rho \cdot L^(2)\cdot h` (1)

Where:

`m` - Mass of the continent, measured in kilograms.

`\rho` - Average density of the rock, measured in kilograms per cubic meter.

`L` - Side of the continent, measured in meters.

`h` - Depth of the continent, measured in meters.

If we know that
`\rho = 2620\,(kg)/(m^(3))`,
`L = 4.450* 10^(6)\,m` and
`h = 31* 10^(3)\,m`, then the mass of the continent is:

`m = \left(2620\,(kg)/(m^(3)) \right)\cdot (4.450* 10^(6)\,m)^(2)\cdot (31* 10^(3)\,m)`

`m = 1.608* 10^(21)\,kg`

The mass of the continent is approximately
`1.608* 10^(21)` kilograms.

2) By assuming that continent can be represented as a particle, we define its kinetic energy as:

`K = (1)/(2)\cdot m \cdot v^(2)` (2)

Where:

`K` - Translational kinetic energy, measured in joules.

`v` - Motion rate of the continent, measured in meters per second.

If we know that
`m = 1.608* 10^(21)\,kg` and
`v = 1* 10^(-2)\,(m)/(s)`, then the kinetic energy of the continent is:

`K = (1)/(2)\cdot (1.608* 10^(21)\,kg)\cdot \left(1* 10^(-2)\,(m)/(s) \right)^(2)`

`K = 8.04* 10^(16)\,J`

The kinetic energy of the continent is approximately
`8.04* 10^(16)` joules.

3) The speed of the jogger is derived from the definition of translational kinetic energy:

`v = \sqrt{(2\cdot K)/(m) }`

If we know that
`K = 8.04* 10^(16)\,J` and
`m = 77\,kg`, then the expected speed of the jogger is:

`v = \sqrt{(2\cdot (8.04* 10^(16)\,J))/(77\,kg) }`

`v\approx 45.698* 10^(6)\,(m)/(s)`

The speed of the 77 kg-jogger would be approximately
`45.698* 10^(6)` meters per second.