Final answer:
The work done by the truck engine while the truck increases its speed from 36 km/h to 72 km/h is the truck's change in kinetic energy, which is calculated to be 150,000 Joules using the kinetic energy formula.
Step-by-step explanation:
The question asks about the work done by a truck engine when a truck changes its speed from 36 km/h to 72 km/h over the course of 2 minutes. Work done by the engine is the change in the kinetic energy of the truck, which can be calculated using the kinetic energy formula:
KE = ½ * mv²
Firstly, we need to convert the speeds from km/h to m/s:
- 36 km/h = 10 m/s
- 72 km/h = 20 m/s
- Next, we can calculate the change in kinetic energy (∆KE) as:
∆KE = ½ * m * (v_final² - v_initial²)
For a truck with a mass (m) of 1000 kg, initial velocity (v_initial) of 10 m/s, and final velocity (v_final) of 20 m/s, the calculation is as follows:
∆KE = 0.5 * 1000 kg * (20 m/s)² - (10 m/s)²
∆KE = 0.5 * 1000 kg * (400 m²/s² - 100 m²/s²)
∆KE = 0.5 * 1000 kg * 300 m²/s²
∆KE = 150,000 J
Hence, the work done by the truck engine is 150,000 Joules.