Answer:
The distance to come to a halt is approximately 49.53 meters
Explanation:
Thee given parameters are;
The speed of the car, v = 50 mph = 22.35 m/s
The mass of the car, m = 1000 kg
The coefficient of friction, = 0.514
The force of friction of the brake = Mass × Gravity × Friction = 1000 × 9.81 × 0.514 = 5042.34 N
The initial kinetic energy of the car = 1/2×m×v² = 1/2 × 1000 × 22.35² = 249761.25 J
The work done by the brake = Force of the brake × Distance, d, to come to halt
By conservation of energy, we have;
The work done by the brake = The initial kinetic energy of the car
∴ The initial kinetic energy of the car = Force of the brake × Distance, d, to come to halt
The initial kinetic energy of the car = 249761.25 J = 5042.34 N × Distance, d, to come to halt
∴The distance to come to a halt = 249761.25 J /(5042.34 N) ≈ 49.53 meters
The distance to come to a halt ≈ 49.53 meters.
The car will cover 49.65 m distance before stopping due to application of brake.
Given data:
The mass of car is, M = 1000 kg.
The initial speed of car is, u = 50.0 mph = (50)(0.447 m/s) = 22.35 m/s.
The coefficient of static friction is, [tex]\mu = 0.514[/tex].
The given problem can be solved using the third kinematic equation of motion. Which is,
[tex]v^{2}=u^{2}+2as[/tex] ....................................................(1)
Here, v is the final speed, v = 0 (Because car is finally stopping)
s is the distance covered before stopping and a is the magnitude of acceleration.
Now, since frictional force opposes the motion. Then,
[tex]F = f\\\\ma = \mu \times m \times g\\\\a = 0.514 \times 9.8\\\\a =5.03 \;\rm m/s^{2}[/tex]
Substituting the values in equation (1) as,
[tex]0^{2}=23.35^{2}+2(-5.03)s\\\\s = \dfrac{2500}{2 \times 5.03}\\\\ s=49.65 \;\rm m[/tex]
Thus, the car will cover 49.65 m distance before stopping due to application of brake.
Learn more about the Kinematic equations of motion here:
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