a woman weighs Fg=650 when standing on a stationary scale. The woman is riding an elevator from the 1st floor to the 10th floor. As the elevator approaches the 10th floor, it decreases its upward speed from 9 m/s to 1 m/s in a time interval of 1 second. What is the average force exerted by the elevator floor on the woman during this 1 second interval? g=10m/s^2 My answer is 260 N. Is this correct?
I figured it out. 130N. I forgot about gravity.
To determine the average force exerted by the elevator floor on the woman during the 1-second interval, we need to consider the change in momentum.
First, let's find the initial momentum of the woman when the elevator starts slowing down. The momentum (p) of an object is calculated by multiplying its mass (m) by its velocity (v):
p = m * v
The mass of the woman (m) is given by her weight divided by the acceleration due to gravity (g):
m = Fg / g
Substituting the given values, we have:
m = 650 N / 10 m/s^2 = 65 kg
Next, we calculate the initial momentum (p_initial) when the elevator is moving at a speed of 9 m/s:
p_initial = m * v_initial = 65 kg * 9 m/s = 585 kg m/s
Now, we determine the final momentum (p_final) when the elevator slows down to a speed of 1 m/s. Since the final momentum is the same as the initial momentum (according to the principle of conservation of momentum), we have:
p_final = p_initial = 585 kg m/s
To find the change in momentum, we subtract the initial momentum from the final momentum:
Δp = p_final - p_initial = 585 kg m/s - 585 kg m/s = 0
Now, we can find the average force (F_avg) exerted by the elevator on the woman using the formula:
F_avg = Δp / Δt
Since Δp is 0 (there is no change in momentum), the average force exerted by the elevator is also 0 N. Hence, your answer of 260 N is not correct.