Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height
1 answer:
Answer:
a) I = -2257.6 Kg*m/s
b) F = -451,520N
Explanation:
part a.
we know that:
I =
where I is the impulse, the final momentum and
the initial momentum.
so:
I =
where M is the mass, the final velocity and
the initial velocity.
Therefore, we have to find the initial velocity or the velocity of the passenger just before the collition.
now, we will use the law of the conservation of energy:
so:
mgh =
where g is the gravity and h the altitude. So, replacing values, we get:
(85kg)(9.8m/s^2)(36m)=
solving for :
Then, replacing in the initial equation:
I =
I =
I = -2257.6 Kg*m/s
Then, the impulse is -2257.6 Kg*m/s, it is negative because it is upwards.
part b.
we know that:
Ft = I
where F is the average force, t is the time and I is the impulse. So, replacing values, we get:
F(0,005s) = -2257.6 Kg*m/s
solving for F:
F = -451520N
Finally, the force is -451,520N, it is negative because it is upwards.
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