a 2390 kg car traveling to the west at 22.4 m/s slows down uniformly under a force of 8740 N to the east.
(a) How much force would be required to cause the same acceleration on a car of mass 3140 kg? Answer in units N
(b) How far would the car move before stopping? Answer in units of m.
(a) If the two cars have the same deceleration rate, the applied force must be proportional to the mass. The speed does not matter for this part.
(b) The deceleration rate is a = F/m = 3.66 m/s^2
The average speed while decelerating is Vav = 22.4 m/s/2 = 11.2 m/s
The time spent decelerating is T = Vo/a = 6.12 seconds
Distance travelled = Vav*T = ?
To find the answers, we can utilize Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, we are given the mass of the car, the initial velocity, the force, and asked to find the acceleration and the distance.
Let's start with part (a):
(a) To find the force required to cause the same acceleration on a car of mass 3140 kg, we can use the formula:
Force = mass * acceleration
We already know the force and the mass, so we need to find the acceleration for the first car.
First, let's calculate the acceleration of the first car using the force applied to it:
Force = mass * acceleration
-8740 N = 2390 kg * acceleration
Now, solving for acceleration:
acceleration = -8740 N / 2390 kg
acceleration ≈ -3.66 m/s²
Now, we can calculate the force required for the car with a mass of 3140 kg:
Force = mass * acceleration
Force = 3140 kg * -3.66 m/s²
Calculating the force:
Force ≈ -11480 N
So, the force required to cause the same acceleration on a car with mass 3140 kg is approximately -11480 N (eastward).
Now, let's move on to part (b):
(b) To find the distance traveled before the car stops, we first need to find the time it takes for the car to stop.
We can use the equation of motion:
velocity² = initial velocity² + 2 * acceleration * distance
Since the final velocity is 0 m/s (the car stops), we have:
0² = (22.4 m/s)² + 2 * acceleration * distance
Rearranging the equation:
2 * acceleration * distance = -22.4 m/s * -22.4 m/s
2 * acceleration * distance = 501.76 m²/s²
Now, substitute the acceleration value we found earlier:
2 * -3.66 m/s² * distance = 501.76 m²/s²
Simplifying further:
-7.32 m/s² * distance = 501.76 m²/s²
Now solve for the distance:
distance = 501.76 m²/s² / -7.32 m/s²
distance ≈ -68.57 m²
Taking the square root of the squared distance, we get:
distance ≈ 8.28 m
So, the car would move approximately 8.28 meters before stopping.