Answer:
a) I = 0 N s, b) v = -3.935 m / s, c) vf = 3.935 m / s, d) y = 0.790 m
Explanation:
a) Let's start by defining the upward direction (+ y) as positive. For this part of the exercise we must use the momentum relationship
I = ∫F dt
The force of the woman on the floor is given and by action and rection the floor exerts on the woman a force of equal magnitude, but opposite direction
I = ∫ (9200 t - 11500 t2) dt
I = 9200 t² / 2 - 11500 t³ / 3
We evaluate between the lower limit t = 0 and upper limit t = 0.800 s
I = 9200 (0.8² -0) - 11500 (0.8³ -0)
I = 5888 -5888
I = 0 N s
Directed from the floor to the woman
b) For this part we use kinematics
v² = v₀² - 2g y
v = √ (0 - 2 9.8 (-0.79))
v = 3.935 m / s
The speed direction is down
c) for this we use the relationship between momentum and the amount of movement
I = ΔP
I = m vf - m v₀
vf = (I + m v₀) / m
This is the impulse of women on the floor
vf = ( 0 + 68 (3.935)) / 68
vf = 3.935 m / s
d) let's use kinematics
v₂ = v₀² - 2gy
0 = v₀² - 2gy
y = v₀² / 2g
y = 3.935²/2 9.8
y = 0.790 m
