Question

In 2006, NASA’s Mars Odyssey orbiter detected violent gas eruptions on Mars, where the acceleration due to gravity is 3.7 m/s2. The jets throw sand and dust about 62.0 m above the surface. Scientists estimate that the jets originate as high-pressure gas speeds through vents just underground at about 130 km/h. How much energy per kilogram of material is lost due to nonconservative forces as the high-speed matter forces its way to the surface and into the air? (Express your answer to two significant figures.)

Answer 1

The unit energy losses due to nonconservative forces is 881.40 joules per kilogram.

Step-by-step explanation:

We can estimate the unit energy losses of gas eruption by Principle of Energy Conservation and Work-Energy Theorem:

`U_(g,1) + K_(1) = U_(g,2)+K_(2)+W_(loss)` (Eq. 1)

Where:

`U_(g,1)` - Gravitational potential energy of gas eruptions at surface, measured in joules.

`U_(g,2)` - Gravitational potential energy of gas eruptions at highest height, measured in joules.

`K_(1)` - Translational kinetic energy of gas eruptions at surface, measured in joules.

`K_(2)` - Translational kinetic energy of gas eruptions at highest height, measured in joules.

`W_(loss)` - Energy losses due to nonconservative forces, measured in joules.

We clear the component associated with energy losses in (Eq. 1):

`W_(loss) = U_(g,1)-U_(g,2)+ K_(1)-K_(2)`

And we expand it afterwards:

`W_(loss) = m\cdot g\cdot (z_(1)-z_(2)) + (1)/(2)\cdot m \cdot (v_(1)^(2)-v_(2)^(2))` (Eq. 2a)

`w_(loss) = g\cdot (z_(1)-z_(2))+(1)/(2)\cdot (v_(1)^(2)-v_(2)^(2))` (Eq. 2b)

Where:

`W_(loss)` - Energy losses due to nonconservative forces, measured in joules.

`w_(loss)` - Unit energy losses due to nonconservative forces, measured in joules per kilogram.

`g` - Gravitational acceleration, measured in meters per second.

`z_(1)`,
`z_(2)` - Bottom and top height, measured in meters.

`v_(1)`,
`v_(2)` - Gas eruption speeds at surface and highest heights, measured in meters per second.

If we know that
`g = 3.7\,(m)/(s^(2))`,
`z_(1) = 0\,m`.
`z_(2) =62\,m`.
`v_(1) = 36.111\,(m)/(s)` and
`v_(2) = 0\,(m)/(s)`, the unit energy losses are:

`w_(loss) = \left(3.7\,(m)/(s^(2)) \right)\cdot (62\,m-0\,m)+(1)/(2) \cdot \left[\left(36.11\,(m)/(s) \right)^(2)-\left(0\,(m)/(s) \right)^(2)\right]`

`w_(loss) = 881.40\,(J)/(kg)`

The unit energy losses due to nonconservative forces is 881.40 joules per kilogram.