Question

Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height of 30.0 m. During the collision at the bottom of the elevator shaft, a 94.0 kg passenger is stopped in 6.00 ms. (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision? If the passenger were to jump upward with a speed of 8.00 m/s relative to the cab floor just before the cab hits the bottom of the shaft, what are the magnitudes of the (c) impulse and (d) average force (assuming the same stopping time)?

Answer 1

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$v_{f}$ - m v₀

Let's calculate the final speed using kinematics, as the cable breaks the initial speed is zero

   $v_{f}$² = V₀² - 2g y

   $v_{f}$² = 0 - 2 9.8 30.0

   $v_{f}$ = √588

   $v_{f}$ = 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 $v_{f}$ - 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