A car traveling 72.0 km/h along a straight, level road is brought uniformly to a stop in a distance of 40.0 m. If the car weighs 8.8 x 103 N, what is the braking force?
(Work done against braking force) = (initial kinetic energy)
F*X = (1/2) M Vo^2
F = M Vo^2/(2 X)
Be sure to convert 72 km/h to 20 m/s before using the formula.
To find the braking force, we can use Newton's Second Law of Motion, which states that force is equal to mass multiplied by acceleration (F = m * a). In this case, the acceleration can be calculated using the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s since the car stops)
u = initial velocity (72.0 km/h = 20.0 m/s)
a = acceleration (unknown)
s = distance (40.0 m)
Rearranging the equation to solve for acceleration:
a = (v^2 - u^2) / (2 * s)
Now, setting the values:
a = (0^2 - 20.0^2) / (2 * 40.0)
a = -400 / 80
a = -5 m/s^2
Since the car is brought to a stop, the acceleration is negative. However, we can consider the magnitude of acceleration since we only need the magnitude of the braking force.
Next, we can calculate the mass of the car using the weight and acceleration due to gravity (9.8 m/s^2):
Weight = mass * gravity
mass = weight / gravity
mass = 8.8 x 10^3 N / 9.8 m/s^2
mass = 897.96 kg (rounding to three significant figures)
Finally, we can calculate the braking force using Newton's Second Law:
F = m * a
F = 897.96 kg * -5 m/s^2
F = -4489.8 N
Therefore, the magnitude of the braking force is 4489.8 N.