A car traveling at 69 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 59 cm (with respect to the road) while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger's upper torso, which has a mass of 41 kg?
12764.2N
To find the magnitude of the force acting on the passenger's upper torso, 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.
We need to find the acceleration of the passenger's upper torso, which can be determined from the change in velocity. Since the passenger starts from a velocity of 69 km/h and comes to a stop, we need to convert this velocity to meters per second.
First, we convert the velocity from kilometers per hour to meters per second. To do this, we can use the conversion factor: 1 kilometer per hour is equal to 0.2778 meters per second.
Velocity = 69 km/h * 0.2778 m/s = 19.17 m/s
Now, we need to find the time it took for the passenger to decelerate from 19.17 m/s to 0 m/s. This can be calculated using the equation:
Vf = Vi + at
Where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and t is the time.
Since Vf = 0 m/s, Vi = 19.17 m/s, and a is the acceleration we want to find, we can rearrange the equation to solve for t:
t = (Vf - Vi) / a
Plugging in the values:
t = (0 - 19.17) / a
Next, we can determine the displacement (change in position) of the passenger's upper torso, which is given as 59 cm. We need to convert this to meters:
Displacement = 59 cm * 0.01 m/cm = 0.59 m
Now we can use the formula for displacement to calculate the acceleration:
Displacement = (Vi * t) + (0.5 * a * t^2)
Since Vi = 19.17 m/s, t is the time calculated above, and displacement is 0.59 m, we can rearrange the equation to solve for a:
a = (2 * (Displacement - (Vi * t))) / t^2
Plugging in the values:
a = (2 * (0.59 - (19.17 * t))) / t^2
Finally, with the acceleration value, we can calculate the force acting on the passenger's upper torso using Newton's second law:
Force = mass * acceleration
Plugging in the values:
Force = 41 kg * a
Now you can calculate the magnitude of the force acting on the passenger's upper torso by substituting the value of 'a' into the equation.
Initial kinetic energy
= (1/2)mv²
Final kinetic energy
= 0
Work done to dissipate this kinetic energy
= F.d
where d = 0.59 m
0.59F = (1/2)41 kg *(69/3.6)²
Solve for F.
(hint: > 10000 N.)