physic (work,energy and power)

`＊雨後彩虹【☆】

2010-11-07 2:09 am

1. In a road test , John drives his car along a straight horizontal road . The car takes 9.3s to accelerate from rest to 100 km h^-1. The total mass of John and his car is 1400 kg .
a.) show that a speed of 100 km h^-1 is approsimately equal to 27.8 ms^-1.
b.) Find the total kinetic energy of John and his car travelling at 100 km h^-1.Hence estimate the average output power of the car when it is accelerating to 100km h^-1 .
c.)A similar road test is conducted on an inclined road . The car now takes 16.2 s to accelerate from rest to 100 km h^-1 along the road . Assume the output power of the car remains unchanged .
i.) Explain why it takes a longer time for the car to accelerate up an inclined road than along a horizontal road .
ii.) Find the increase in height of the car after accelersting for 16.2 s along the inclined road .
d.) On a certain day , the car was involved in a traffic accident . John braked hard to stop the car and skid marks were left on a horizontal road .Investigation by the police recealed the following information:
lenght of the skid marks = 30.5m
average frictional force between the tyres of the car and the road suface =11200N
i .) Describe the energy change involved when the car was braking .
ii.) John claimed that he was driving at a speed below 70 km h^-1 before the accident . Explain whether John was telling the truth or not .

回答 (1)

天同

2010-11-07 8:11 am

✔ 最佳答案

1. (a) 100 km/h = 100 x 1000 m/3600 s = 27.8 m/s(b) Total kinetic energy = (1/2) x 1400 x 27.8^2 J = 5.41 x 10^5 JPower = 5.41x10^5/9.3 w = 5.82x10^4 w = 58.2 kw(c) (i) When the car moves up an incline, it gains potential energy in addition to kinetic energy. Since the power remains the same as before, a longer time is needed in order to produce more energy.

(ii) Energy produced by the car engine = 5.82x10^4 x 16.2 J = 9.43x10^5 JSince kinetic energy of car = 5.41x10^5 J
hence, gain in potential energy = ( 9.43x10^5 - 5.41x10^5) J = 4.01 x 10^5 J
Increase in height = 4.01x10^5/1400g m = 28.6 m[where g is the acceleration due to gravity, taken to be 10 m/s2]
(d) (i) kinetic energy of car was transformed into heat energy by friction.(ii) Use: work-done = change of kinetic energy11200 x 30.5 = (1/2) x 1400u^2where u is the speed of the car before braking
solve for u gives u = 22.1 m/s = 79.5 km/h
Thus, John was not telling the truth.

👍 0 👎 0