A(n) 1949.1 kg car is coasting along a level road at 22.9 m/s. A constant braking force is applied, such that the car is stopped in a distance of 74.6 m.
What is the magnitude of the braking force? Answer in units of N
See previous post.
To calculate the magnitude of the braking force, we can use the equation:
F = (m * v^2) / (2 * d)
where F is the magnitude of the braking force, m is the mass of the car, v is the initial velocity of the car, and d is the distance over which the car is stopped.
Given:
m = 1949.1 kg (mass of the car)
v = 22.9 m/s (initial velocity of the car)
d = 74.6 m (distance over which the car is stopped)
Plugging in the values into the equation, we get:
F = (1949.1 kg * (22.9 m/s)^2) / (2 * 74.6 m)
Calculating this:
F = (1949.1 kg * 526.41 m^2/s^2) / (2 * 74.6 m)
F = (1024167.149 kg·m^2/s^2) / (149.2 m)
Simplifying:
F ≈ 6862.8 N
Therefore, the magnitude of the braking force is approximately 6862.8 N.
To find the magnitude of the braking force, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
First, we need to find the acceleration of the car. We can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, since the car is brought to a stop)
u = initial velocity (22.9 m/s)
a = acceleration
s = distance (74.6 m)
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (0^2 - (22.9 m/s)^2) / (2 * 74.6 m)
Calculating the acceleration:
a = (-22.9 m/s)^2 / (2 * 74.6 m)
a ≈ -7.029 m/s^2 (the negative sign indicates deceleration)
Now, we can find the magnitude of the braking force using Newton's second law:
F = m * a
Substituting the given mass:
F = 1949.1 kg * -7.029 m/s^2
Calculating the braking force:
F ≈ -13,679 N
Since the question asks for the magnitude of the braking force, we can take the absolute value of the result:
The magnitude of the braking force is approximately 13,679 N.