Two masses, each having a value of M, are vibrating vertically on a spring with a Hooke's law constant, k. At the lowest point o
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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
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
J = 2.5 × 10³Ns
b) Average force is given by
F(avg) = J / Δt
average force on the passenger during the collision = 8.3 × 10⁵ N