What is the magnitude of the average force required to stop an 1400kg car in 9.0s if the car is traveling at 85km/h ?
Force(time)=masscar*changevelociyt
change km/hr to m/s, then solve for force.
To find the magnitude of the average force required to stop the car, we can use Newton's second law of motion, which states that the force applied to an object is equal to its mass multiplied by its acceleration:
F = m * a
First, let's convert the mass of the car from kg to g:
m = 1400 kg = 1400 * 1000 g = 1,400,000 g
Next, we need to calculate the deceleration of the car. We can use the equation of motion:
v = u + a * t
where:
v = final velocity (0 m/s, as the car is coming to a stop)
u = initial velocity (85 km/h)
a = acceleration (negative, as the car is decelerating)
t = time (9.0 s)
Converting the initial velocity from km/h to m/s:
u = 85 km/h * 1000 m/km * 1 h/3600 s = 85 * 1000 / 3600 m/s ≈ 23.6 m/s
Rearranging the equation of motion, we get:
a = (v - u) / t
Plugging in the values and solving for acceleration:
a = (0 - 23.6 m/s) / 9.0 s ≈ -2.6 m/s^2
Now we can substitute the mass and acceleration into Newton's second law to find the force:
F = m * a
≈ 1,400,000 g * -2.6 m/s^2
Finally, we need to convert the force from g·m/s^2 to kg·m/s^2 (N):
1 N = 1000 g·m/s^2
F ≈ (1,400,000 * -2.6) / 1000 kg·m/s^2
≈ -3640 N (Note: the negative sign indicates the force is acting in the opposite direction of the car's motion)
Therefore, the magnitude of the average force required to stop the car is approximately 3640 Newtons.