A 14,245N auto with a velocity of 6.0m/s crashes into a brick wall. If the car moves 64.0 cm before coming to a stop, what average force does the car exert on the wall?
Force x distance = Energy absorbed by wall
= (1/2) M V^2
Solve for Force
To find the average force exerted by the car on the wall, we can use Newton's second law of motion, which states that the force exerted on an object is equal to the object's mass multiplied by its acceleration:
F = m * a
In this case, the car's mass (m) is not given, but we can calculate it using the relationship between force and acceleration. The net force acting on the car is equal to the product of its mass and acceleration:
F = m * a
The acceleration (a) can be calculated using the initial velocity, final velocity, and the distance traveled:
v^2 = u^2 + 2as
Where:
v = final velocity = 0 (since the car comes to a stop)
u = initial velocity = 6.0 m/s
s = distance traveled = 64.0 cm = 0.64 m
Let's solve for 'a':
0 = (6.0)^2 + 2a(0.64)
0 = 36 + 1.28a
a = -36/1.28
a ≈ -28.13 m/s^2 (negative because the car is decelerating)
Now that we have the acceleration, we can substitute it back into the equation F = m * a:
14,245 N = m * (-28.13 m/s^2)
Solving for 'm':
m = 14,245 N / (-28.13 m/s^2)
m ≈ -507 kg
Since mass cannot be negative, we can ignore the negative sign. Therefore, the mass of the car is approximately 507 kg.
Now we can substitute the mass and acceleration back into the force equation to find the average force exerted by the car:
F = m * a
F = 507 kg * (-28.13 m/s^2)
F ≈ -14,256.91 N
Again, since force cannot be negative, we ignore the negative sign. Therefore, the average force exerted by the car on the wall is approximately 14,256.91 N.