A car traveling at 62 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 58 cm (with respect to the road) while being brought to rest by an inflated air bag. What magnitude of force in newtons (assumed constant) acts on the passenger's upper torso, which has a mass of 41 kg?
To find the magnitude of force acting on the passenger's upper torso, we can use Newton's second law of motion, which states that the force on an object is equal to the mass of the object multiplied by its acceleration.
First, let's convert the velocity of the car from km/h to m/s, since the SI unit of force is Newtons (N) and the SI unit of velocity is meters per second (m/s).
Given:
Velocity of the car (v) = 62 km/h
Distance moved by the passenger (d) = 58 cm
Mass of the passenger (m) = 41 kg
1 km = 1000 m, 1 hour = 3600 seconds
So, converting the velocity:
v = 62 km/h = (62 * 1000) / (3600) m/s ≈ 17.22 m/s
Now, we can determine the acceleration of the passenger using the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s in this case, as the passenger comes to rest)
u = initial velocity (17.22 m/s, since the passenger was traveling with the car)
a = acceleration (unknown)
s = distance moved by the passenger (58 cm = 0.58 m)
0^2 = (17.22)^2 + 2a * 0.58
0 = 296.5284 + 1.16a
1.16a = -296.5284
a = -296.5284 / 1.16 ≈ -255.0523 m/s^2
We've obtained the acceleration as a negative value because it acts opposite to the direction of motion.
Now we can calculate the force acting on the passenger using Newton's second law of motion:
F = m * a
F = 41 kg * (-255.0523 m/s^2)
F ≈ -10,468.14 N
Since force is a vector quantity, it has both magnitude and direction. In this calculation, the negative sign indicates that the force is acting in the opposite direction to the motion of the passenger's upper torso.
Therefore, the magnitude of the force acting on the passenger's upper torso is approximately 10,468.14 Newtons (N).
To find the magnitude of force acting on the passenger's upper torso, we can use the equation:
Force = (mass) x (acceleration)
First, we need to find the acceleration of the passenger. We can use the equation of motion:
Final velocity^2 = Initial velocity^2 + 2 x acceleration x displacement
The final velocity is 0 m/s since the passenger comes to rest. The initial velocity can be found by converting the car's speed from km/h to m/s:
Initial velocity = 62 km/h = 62 x 1000 / 3600 m/s = 17.22 m/s
Plugging in these values into the equation, we get:
0^2 = 17.22^2 + 2 x acceleration x 0.58 m
0 = 296.8184 + 1.16 x acceleration
Simplifying the equation, we have:
1.16 x acceleration = -296.8184
acceleration = -255.935 m/s^2
Now, we can calculate the magnitude of force:
Force = (mass) x (acceleration)
Force = 41 kg x -255.935 m/s^2
Force ≈ -10,503.735 N
The magnitude of force acting on the passenger's upper torso is approximately 10,503.735 N. Note that the negative sign indicates the direction of the force (opposite to the direction of the passenger's motion).