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 39.0 m. During the collision at the bottom of the elevator shaft, a 92.0 kg passenger is stopped in 3.00 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

Answer 1

Answer:

a) impulse = 2.5 × 10³Ns

b) average force on the passenger during the collision = 8.3 × 10⁵ N

Explanation:

Given that

Elevator cab free fall = 39.0m

The passenger = 92kg

Collision time = 3.00ms

The impulse is given by

J = Δp

$= m(v_f - v_i)$

where m is the mass ,

v(f) is final velocity

v(i) is initial velocity which is equal to 0

from the conservation energy system, the kinectic enegy of the passenger equal the potential energy of the pasenger

$\frac{1}{2} mV^2_f = mgh$

$V_f = \sqrt{2gh} \\ = \sqrt{2 \times 9.8 \times 39 }\\ = 27.65ms\\impulse = 92 \times 27.65\\= 2.5 \times 10^3Ns$

J = 2.5 × 10³Ns

b) Average force is given by

F(avg) = J / Δt

$F_a_v_g = \frac{2,5 \times 10^3}{3 \times 10 ^-^3} \\8.3 \times 10^5N$

average force on the passenger during the collision = 8.3 × 10⁵ N