Question

An airplane whose mass is \(5000 \mathrm{~kg}\) is flying with a velocity of \(150 \mathrm{~m} / \mathrm{s}\) at an altitude of \(10,000 \mathrm{~m}\), both measured relative to the surface of the earth. The acceleration of gravity can be taken as constant at \(g=9.78 \mathrm{~m} / \mathrm{s}^{2}\). (a) Calculate the kinetic and potential energies of the airplane, both in \(\mathrm{kJ}\). (b) If the kinetic energy increased by \(10,000 \mathrm{~kJ}\) with no change in elevation, what would be the final velocity, in \(\mathrm{m} / \mathrm{s}\) ?

Short answer

KE = 56250 \, \mathrm{kJ}, PE = 489000 \, \mathrm{kJ}, v_{new} \approx 162.80 \, \mathrm{m/s}

Step-by-step solution

Calculate Kinetic Energy

The kinetic energy (KE) of an object is given by the formula \[ KE = \frac{1}{2} m v^2 \] where \( m \) is the mass of the airplane \( v \) is the velocity of the airplane Plug in the given values: \( m = 5000 \, \mathrm{kg} \) \( v = 150 \, \mathrm{m/s} \)\[ KE = \frac{1}{2} \times 5000 \, \mathrm{kg} \times (150 \, \mathrm{m/s})^2 \]Calculate the kinetic energy:\[ KE = 56250000 \, \mathrm{J} \]Convert to \,\mathrm{kJ} (1 \, \mathrm{kJ} = 1000 \, \mathrm{J})\[ KE = 56250 \, \mathrm{kJ} \]

Calculate Potential Energy

The potential energy (PE) of an object at height \(h\) is given by the formula \[ PE = mgh \] where \( m \) is the mass of the airplane \( g \) is the acceleration due to gravity \( h \) is the altitude Plug in the given values: \( m = 5000 \, \mathrm{kg} \) \( g = 9.78 \, \mathrm{m/s^2} \) \( h = 10000 \, \mathrm{m} \)\[ PE = 5000 \, \mathrm{kg} \times 9.78 \, \mathrm{m/s^2} \times 10000 \, \mathrm{m} \]Calculate the potential energy:\[ PE = 489000000 \, \mathrm{J} \]Convert to \,\mathrm{kJ} (1 \, \mathrm{kJ} = 1000 \, \mathrm{J})\[ PE = 489000 \, \mathrm{kJ} \]

Determine Final Velocity with Increased Kinetic Energy

The new KE after the increase is: \[ KE_{new} = 56250 \, \mathrm{kJ} + 10000 \, \mathrm{kJ} = 66250 \, \mathrm{kJ} \]Convert to \,\mathrm{J} (1 \, \mathrm{kJ} = 1000 \, \mathrm{J})\[ KE_{new} = 66250000 \, \mathrm{J} \]The formula for kinetic energy is: \[ KE = \frac{1}{2} m v^2 \]Solve for the new velocity \( v_{new} \)\[ 66250000 = \frac{1}{2} \times 5000 \times v_{new}^2 \]\[ 66250000 = 2500 v_{new}^2 \]\[ v_{new}^2 = \frac{66250000}{2500} \]\[ v_{new}^2 = 26500 \]\[ v_{new} = \sqrt{26500} \]\[ v_{new} \approx 162.80 \, \mathrm{m/s} \]

Key concepts

kinetic energy

Kinetic energy (KE) is the energy possessed by an object due to its motion. The faster something moves, the more kinetic energy it has. This energy is calculated using the formula \( KE = \frac{1}{2} m v^2 \), where:

• \( m \) is the mass of the object in kilograms (kg)
• \( v \) is the velocity (speed) of the object in meters per second (m/s)

For the given airplane, with mass \( m = 5000 \mathrm{~kg} \) and velocity \( v = 150 \mathrm{~m/s} \), we plug these numbers in:
\( KE = \frac{1}{2} \times 5000 \mathrm{~kg} \times (150 \mathrm{~m/s})^2 = 56250000 \mathrm{~J} \).
To convert joules to kilojoules (kJ):
\( 1 \mathrm{~kJ} = 1000 \mathrm{~J} \)
Hence, \( KE = 56250 \mathrm{~kJ} \).
This energy is strictly dependent on the object's speed and mass.

potential energy

Potential energy (PE) refers to the energy stored in an object because of its position or height. For objects at height, gravitational potential energy is considered, and it is calculated using the formula:
\( PE = mgh \), where:

• \( m \) is the mass in kilograms (kg)
• \( g \) is the acceleration due to gravity (\(9.78 \mathrm{~m/s^2} \))
• \( h \) is the height above the reference point (in meters)

For the airplane at an altitude \( h = 10000 \mathrm{~m} \), the potential energy calculation is:
\( PE = 5000 \mathrm{~kg} \times 9.78 \mathrm{~m/s^2} \times 10000 \mathrm{~m} = 489000000 \mathrm{~J} \).
Converted to kilojoules, this is \( 489000 \mathrm{~kJ} \).
Unlike kinetic energy, potential energy is dependent on the object's vertical position relative to the ground.

velocity calculation

Velocity calculation comes into play when determining the change in kinetic energy. Suppose kinetic energy increases by \(10000 \mathrm{~kJ} \) without any change in height. The new kinetic energy becomes:
\( KE_{new} = 56250 \mathrm{~kJ} + 10000 \mathrm{~kJ} = 66250 \mathrm{~kJ} \).
Converting this back to joules: \( KE_{new} = 66250000 \mathrm{~J} \).
Using the kinetic energy formula \( KE = \frac{1}{2} m v^2 \), we solve for the new velocity (\( v_{new} \)):
\( 66250000 = \frac{1}{2} \times 5000 \times v_{new}^2 \).
Simplifying,
\( 66250000 = 2500 \times v_{new}^2 \)
\( v_{new}^2 = \frac{66250000}{2500} = 26500 \)
\( v_{new} = \sqrt{26500} \approx 162.80 \mathrm{~m/s} \).
This shows how increasing kinetic energy influences velocity, following the principles of energy conservation.