What net braking force must be applied to stop a car with a mass of 900 kg initially travelling at a velocity of 100 km/h within a straight-line distance of 50 m?
(Braking Force) x distance = Initial Kinetic Energy
Force times distance is work done against friction, which must equal the initial kinetic energy.
Change 100 km/h to m/s before calculating Kinetic Energy
For the Kinetic Energy formula, read my answer to your previous question.
200
To solve this problem, we can use the principle of energy conservation and the equations of motion.
First, let's convert the initial velocity of the car from km/h to m/s. We know that 1 km = 1000 m and 1 hour = 3600 seconds. Therefore:
Initial velocity (V) = 100 km/h = (100 * 1000 m) / (3600 s) = 27.78 m/s
Next, we need to find the final velocity (Vf). Since the car comes to a stop, the final velocity will be 0 m/s.
Now, we can use the equation of motion to find the acceleration (a) of the car:
Final velocity (Vf) = Initial velocity (V) + (acceleration (a) * time (t))
Since the car comes to a stop, Vf = 0 m/s. Therefore:
0 = 27.78 m/s + (a * t)
The time taken (t) can be found using the equation:
Distance (d) = Initial velocity (V) * time (t) + 0.5 * acceleration (a) * time (t)^2
In this case, the distance (d) is given as 50 m, the initial velocity (V) is 27.78 m/s, and the acceleration (a) is what we need to find.
Plugging in the values, we get:
50 m = 27.78 m/s * t + 0.5 * a * t^2
Now, we have two equations with two unknowns (a and t). We can solve for t in the second equation:
t = (sqrt((V^2) + 2ad) - V) / a
Substituting this value of t into the first equation:
0 = 27.78 m/s + (a * ((sqrt((V^2) + 2ad) - V) / a))
Simplifying this equation:
0 = 27.78 m/s + sqrt((V^2) + 2ad) - V
Rearranging the equation:
sqrt((V^2) + 2ad) = V - 27.78 m/s
Squaring both sides:
(V^2) + 2ad = (V - 27.78 m/s)^2
Expanding and simplifying:
(V^2) + 2ad = V^2 - (2 * 27.78 m/s * V) + (27.78 m/s)^2
Substituting the known values, we get:
2a * 50 m = -(2 * 27.78 m/s * 27.78 m/s)
Simplifying:
100a = -1544.2812
Finally, solving for acceleration (a):
a = -1544.2812 / 100
a ≈ -15.44 m/s²
Since we are interested in the net braking force (F), we can use Newton's second law of motion:
F = m * a
Substituting the known values, we get:
F = 900 kg * (-15.44 m/s²)
F ≈ -13,896 N (negative sign indicates that the force is in the opposite direction of motion)
Therefore, approximately a net braking force of -13,896 N (or 13,896 N in the direction opposite to the car's motion) must be applied to stop the car.