What will be the average braking force needed to bring a car of mass 800kg traveling 72km/hr to a stop at a distance of 100m
F = m*a
First, we find the acceleration.
Use the following equation of motion:
2*a*s = v^2 - u^2
s = 100m, v = 0, u = 72km/hr
= 20m/s
=> 2*a*100 = 0 - (20)^2
=> a = -400/200
= -2m/s^2
The sign is negative because the acceleration is retarding in nature. Similarly, the force will also have a negative sign:
F = m*a
= 800*(-2)
= -1600N
Hence, the magnitude of the force required will be 1600N
and, there is another method.
starting velocity=20m/s, ending velocity=0, average velocty=10m/s
time to go 100m, time=distance/avgV=10sec
Vf=vi+at
0=20)a*10
a=-2m/s then us F=ma to get force.
To find the average braking force needed to bring a car to a stop, we can use the formula:
F = (m * v^2) / (2 * d),
where F is the braking force, m is the mass of the car, v is the initial velocity, and d is the distance traveled.
First, let's convert the velocity from km/hr to m/s. We can do this by dividing by 3.6:
v = 72 km/hr / 3.6 = 20 m/s.
Now, substituting the given values into the formula:
F = (800 kg * (20 m/s)^2) / (2 * 100 m).
Simplifying further:
F = (800 kg * 400 m^2/s^2) / 200 m = 160,000 N / 200 m.
F = 800 N.
Therefore, the average braking force needed to bring the car to a stop is 800 Newtons.