What force is required to stop a 1500-kg car in a distance of 0.20 m if it is initially moving at 2.2 m/s?
V^2 = Vo^2 + 2a*d.
0 = (2.2)^2 + 0.4a,
a = -12.1 m/s^2.
F = M*a = 1500*(-12.1) = -18,150 Joules.
The negative sign means the force opposes the motion.
work done by force = F d = 0.20 d Joules
kinetic energy = (1/2) m v^2
work done to stop = kinetic energy
0.20 d = (1/2)(1500)(2.2)^2
banda
To determine the force required to stop a car, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a), or F = ma.
In this case, we need to find the acceleration of the car in order to determine the force required to stop it. We can use the equation of motion to calculate the acceleration: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
Given:
- Initial velocity (u) = 2.2 m/s
- Distance traveled (s) = 0.20 m
- Final velocity (v) = 0 m/s (since the car needs to come to a stop)
Rearranging the equation of motion to solve for acceleration:
v^2 = u^2 + 2as
0^2 = 2.2^2 + 2a(0.20)
0 = 4.84 + 0.4a
-4.84 = 0.4a
a ≈ -12.1 m/s^2
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, which is necessary to bring the car to a stop.
Now, we can use Newton's second law to find the force required:
F = ma
F = 1500 kg × (-12.1 m/s^2)
F ≈ -18,150 N
The negative sign in the force indicates that the force is applied opposite to the direction of motion, as required to stop the car. Therefore, the force required to stop the 1500-kg car in a distance of 0.20 m, when initially moving at 2.2 m/s, is approximately 18,150 newtons.