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
[tex]= m(v_f - v_i)[/tex]
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
[tex]\frac{1}{2} mV^2_f = mgh[/tex]
[tex]V_f = \sqrt{2gh} \\ = \sqrt{2 \times 9.8 \times 39 }\\ = 27.65ms\\impulse = 92 \times 27.65\\= 2.5 \times 10^3Ns[/tex]
J = 2.5 × 10³Ns
b) Average force is given by
F(avg) = J / Δt
[tex]F_a_v_g = \frac{2,5 \times 10^3}{3 \times 10 ^-^3} \\8.3 \times 10^5N[/tex]
average force on the passenger during the collision = 8.3 × 10⁵ N
