A 14,700 Newton car is traveling at 25 m/s. The brakes are applied suddenly, and the car slides to a stop. The average breaking force between the tires and road is 7100 Newtons. How far will the car slide once the brakes are applied?
2006-12-03 13:28:09 · 3 answers · asked by josh h 3 in Science & Mathematics ➔ Physics
This is a question of energy dissipation. The mass of the car is given by 14700/9.8 kg = 1500 kg. The inital kinetic energy of the car is 0.5 * 1500 kg * (25 m/s)^2 = 468750J
The brakes must do this amount of work to stop the car, and since work is Force * distance, or W = F*d,
Solving for d,
d = W/F
The car stops the car in
d = 468750/7100 m = 66.02. m
2006-12-03 13:37:17 · answer #1 · answered by Anonymous · 0⤊ 0⤋
You must be wondering what this newton is doing here to tell you about car. But it is just another way of telling you the car's mass. Its mass is such that earth attracts it with a force of 14,770 N. So its mass, m is 14700/9.8 = 1550 kg. Now the opposing force applied on the car is 7100 N. This produces deceleration, a in the car = 7100/1500 = 4.74 m/s^2. Use v^2 - u^2 = 2as formula to calculate s
2006-12-03 21:49:41 · answer #2 · answered by Let'slearntothink 7 · 0⤊ 0⤋
m=14,700 / g =about 1,470 kg
a=7100N/1,470 kg=4.8 m/s^2
t=25/4.8=5.2s to stop
so distance = x=0.5*a*t^2
x=0.5*4.8*5.2^2=65m
2006-12-03 21:35:12 · answer #3 · answered by Nick C 4 · 0⤊ 0⤋