A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ft/s along
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Answer:
a) I = 2279.5 N s , b) F = 3.80 10⁵ N, c) I = 3125.5 N s and d) F = 5.21 10⁵ N
Explanation:
The impulse is equal to the variation in the amount of movement.
I =∫ F dt = Δp
I = m - m v₀
Let's calculate the final speed using kinematics, as the cable breaks the initial speed is zero
² = V₀² - 2g y
² = 0 - 2 9.8 30.0
= √588
= 24.25 m/s
a) We calculate the impulse
I = 94 24.25 - 0
I = 2279.5 N s
b) Let's join the other expression of the impulse to calculate the average force
I = F t
F = I / t
F = 2279.5 / 6 10⁻³
F = 3.80 10⁵ N
just before the crash the passenger jumps up with v = 8 m / s, let's take the moments of interest just when the elevator arrives with a speed of 24.25m/s down and as an end point the jump up to vf = 8 m / n
c) I = m - m v₀
I = 94 8 - 94 (-24.25)
I = 3125.5 N s
d) F = I / t
F = 3125.5 / 6 10⁻³
F = 5.21 10⁵ N