In an Automobile a 44 kg passenger moving at 40 m/s is brought to rest by an air bag during a 0.1 s what is the magnitude of the average force exerted on the passenger during this time ?
force*time=mass*changevelocity
force=44*40/.1 N
To find the magnitude of the average force exerted on the passenger, we can use 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.
First, we need to find the acceleration of the passenger. We can use the equation:
acceleration = (final velocity - initial velocity) / time
Given:
Mass of the passenger (m) = 44 kg
Initial velocity (u) = 40 m/s
Time (t) = 0.1 s
Substituting the values into the equation, we get:
acceleration = (0 - 40) / 0.1
acceleration = -400 m/s^2
Since the passenger is brought to rest, the final velocity is zero, and the acceleration is negative.
Next, we can calculate the force using Newton's Second Law:
force = mass * acceleration
Substituting the values, we have:
force = 44 kg * -400 m/s^2
force = -17600 N
The magnitude of the force is always positive, so we take the absolute value:
magnitude of force = |-17600 N|
magnitude of force = 17600 N
Therefore, the magnitude of the average force exerted on the passenger during this time is 17600 Newtons.