gravitational potential energy

Question

In magazine car tests an important indicator of performance is the zero to 65 mph (0 to 104.6 km/h) acceleration time. A time below 6 s is considered to be extremely quick. If a car with mass 1300 kg were dropped vertically from a great height, how long would it take to go from zero to 60 mph and what would the magnitude of the change in gravitational potential energy be during this time? (neglect air resistance)

Answer 1

v = a·t, so v/a = t

• v = velocity of the car = 104.6 kph
• a = acceleration due to gravity = 9.8 m/s
• t = time

 

Velocity is given in km/hr.  First convert it to m/s.

 

(104.6 km/hr)(1000 m/km)(1/3600 hr/sec) = 29.1 m/s

 

v/a = t

(29.1)/(9.8) ≈ 3 sec

 

The distance the car falls in 3 seconds is:

 

h = (1/2)at²

h ^{= (}1/2)(-9.8)(3²)

 

Use your calculator to get the answer.  The change in potential energy is mgh

• m = car's mass = 1300 kg
• g = acceleration due to gravity = -9.8 m/s
• h = the distance you computed above

 

Plug the numbers into the formula and compute the answer.