Final answer:
To determine the minimum stopping distance needed to avoid injury, you need to calculate the force exerted by the shock-absorbing device on the passenger. The force exerted is equal to the mass of the passenger multiplied by the deceleration experienced by the elevator. By rearranging the equations and substituting the given values, we can calculate that the minimum stopping distance is approximately 0.176 meters.
Step-by-step explanation:
To determine the minimum stopping distance needed to avoid injury, we need to calculate the force exerted by the shock-absorbing device on the upright body of the passenger. The force exerted by the floor of the elevator is equal to the mass of the passenger multiplied by the acceleration due to gravity. According to Newton's second law, this force is also equal to the mass of the passenger multiplied by the deceleration experienced by the elevator:
F = m * a
Where F is the force, m is the mass, and a is the deceleration. Rearranging this equation, we can solve for a:
a = F / m
Now we can substitute the given values: F = 8000 N and m = 65 kg:
a = 8000 N / 65 kg
a ≈ 123.08 m/s^2.
The deceleration experienced by the elevator is equal to the acceleration due to gravity minus the acceleration needed to stop the elevator:
a = g - a_stop
Where g is the acceleration due to gravity, approximately 9.8 m/s^2. Rearranging this equation, we can solve for a_stop:
a_stop = g - a
a_stop = 9.8 m/s^2 - 123.08 m/s^2
a_stop ≈ -113.28 m/s^2.
Since the deceleration is negative, it indicates that the elevator is decelerating in the opposite direction of its initial motion. To calculate the minimum stopping distance, we can use the equation:
d = v^2 / (2*a_stop)
Where d is the distance, v is the initial velocity, and a_stop is the deceleration. Substituting the given values: v = 20 m/s and a_stop = -113.28 m/s^2:
d = (20 m/s)^2 / (2 * (-113.28 m/s^2))
d ≈ 0.176 m.
Therefore, the minimum stopping distance needed to avoid injury is approximately 0.176 meters.