(a) What is the kinetic energy in joules of a 960-kg automobile traveling at 70 km/h? J (b) How much work would have to be done to bring a 960-kg automobile traveling at 70 km/h to a stop? J

Answer 1

(a) What is the kinetic energy in joules of a 960-kg automobile traveling at 70 km/h?

First convert the velocity, 70 km/h, to m/s by the use of appropriate conversion factors:

v = 70 km/h * (1000 m/ 1 km) * (1 h / 3600 s)

KE = 1/2 m v² = 1/2 (960 kg) (70 km/h * (1000 m/ 1 km) * (1 h / 3600 s))² = 181481.481481 J

With two significant digits this would round to just 180 000 J. Of course more or less digits may be considered significant.

(b) How much work would have to be done to bring a 960-kg automobile traveling at 70 km/h to a stop?

The work-energy theorem gives that the work done will equal the change in KE. At a stop the KE will be 0, as at a stop the v = 0. So The change in KE, will be equal to the initial KE, calculated above in part (a). In short the answer for (b) is the same as the answer for (a), 180 000 J.

I hope that helps! Let me know if I can clarify any points.