Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson,

*Physics, 2/e*, Problems 6.71, 6.75

A 2013 Chevrolet Camaro ZL1 (mass 1,984 kg) on a horizontal track takes 8.8 s to accelerate from 13 m/s to 22 m/s, and then takes 7.7 s to accelerate from 22 m/s to 31 m/s[*]. Which process used a greater average mechanical power output from the engine?

(A) Accelerating from 13 m/s to 22 m/s.

(B) Accelerating from 22 m/s to 31 m/s.

(C) (There is a tie.)

(D) (Not enough information is given.)

[*] "Curb weight: 4,373 lb; Top gear, 30–50 mph: 8.8 sec; Top gear, 50–70 mph: 7.7 sec," caranddriver.com/reviews/2013-chevrolet-camaro-zl1-convertible-test-review.

Correct answer (highlight to unhide): (B)

The mechanical output of the car's engines does non-conservative work to increase translational kinetic energy:

*W*= ∆

_{nc}*KE*,

_{tr}where the average power output of the car's engines is the rate of work done per time:

*P*=

_{av}*W*/∆

_{nc}*t*= ∆

*KE*/∆

_{tr}*t*.

For accelerating from 13 m/s to 22 m/s, the engine power output is:

*P*= (1,984 kg)·((22 m/s)

_{av}^{2}– (13 m/s)

^{2})/(8.8 s) = 3.5×10

^{4}watts,

and for accelerating from 22 m/s to 31 m/s, the engine power output is:

*P*= (1,984 kg)·((31 m/s)

_{av}^{2}– (22 m/s)

^{2})/(7.7 s) = 6.1×10

^{4}watts.

Thus accelerating from 22 m/s to 31 m/s in 7.7 s requires more power output for this car than accelerating from 13 m/s to 22 m/s in 8.8 s.

Sections 70854, 70855

Exam code: midterm02gL0u

(A) : 7 students

(B) : 45 students

(C) : 2 students

(D) : 0 students

Success level: 83%

Discrimination index (Aubrecht & Aubrecht, 1983): 0.33