To understand what happens to the plane's kinetic energy on the two flights, we need to consider the relevant factors that contribute to kinetic energy.
Kinetic energy (KE) is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where:
- \(m\) is the mass of the object (in this case, the airplane and the passengers),
- \(v\) is the velocity of the object.
-
Initial Clubbed Mass: The kinetic energy will depend on both the mass of the airplane and the total mass of the passengers being carried.
- On the flight from Phoenix to Los Angeles, the plane carries 320 passengers.
- On the return flight to Phoenix, the plane carries 264 passengers.
-
Kinetic Energy Calculations:
-
For the flight from Phoenix to Los Angeles:
- Let \(m_1\) be the mass of the airplane plus the mass of 320 passengers.
- The velocity \(v\) is 490 mph.
- Thus, the kinetic energy for this flight would be: \[ KE_1 = \frac{1}{2} m_1 (490)^2 \]
-
For the return flight from Los Angeles to Phoenix:
- Let \(m_2\) be the mass of the airplane plus the mass of 264 passengers.
- The velocity \(v\) remains 490 mph.
- So, the kinetic energy for this flight would be: \[ KE_2 = \frac{1}{2} m_2 (490)^2 \]
-
-
Comparison of Mass: Since \(m_1\) (mass with 320 passengers) is greater than \(m_2\) (mass with 264 passengers), we can state that:
- \(m_1 > m_2\)
-
Change in Kinetic Energy:
- Since the airplane's speed remains constant at 490 mph, the difference in kinetic energy between the two flights depends solely on the differences in mass. Specifically, the kinetic energy on the outbound flight \(KE_1\) (with more passengers) is greater than the kinetic energy on the return flight \(KE_2\) (with fewer passengers).
In conclusion, the plane's kinetic energy decreases on the return flight from Los Angeles to Phoenix due to a reduction in the total mass carried (fewer passengers).
1 answer