Final answer:
The average power required by a truck to increase the velocity of a 2000.0 kg object from rest to 25.0 m/s in 120.0 seconds is 5208.3 W.
Step-by-step explanation:
A student asked how much average power is required for a truck to tow an object with a mass of 2000.0 kg from rest to a velocity of 25.0 m/s in a time interval of 120.0 s. To find the average power, we first need to calculate the work done by the truck on the object using the work-energy principle.
Work done (W) is equal to the change in kinetic energy, which is:
W = \( \frac{1}{2} \) m(v2 - u2)
where m is mass, v is the final velocity, and u is the initial velocity. Since the initial velocity u is 0 (object starts from rest), the work done is:
W = \( \frac{1}{2} \) \(2000.0 kg\) \(25.0 m/s\)2
W = 625,000 J
To find average power P, we divide the work done by the time taken, t:
P = \( \frac{W}{t} = \frac{625,000 J}{120.0 s} \)
P = 5208.3 W
Therefore, the correct answer is a. 5208.3 W.