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Tju [1.3M]
3 years ago
14

Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height

of 36.0 m. During the collision at the bottom of the elevator shaft, a 85.0 kg passenger is stopped in 5.00 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
Physics
1 answer:
butalik [34]3 years ago
3 0

Answer:

a) I = -2257.6 Kg*m/s

b) F = -451,520N

Explanation:

part a.

we know that:

I = P_f-P_i

where I is the impulse, P_f the final momentum and P_i the initial momentum.

so:

I = MV_f-MV_i

where M is the mass, V_f the final velocity and V_i 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:

E_i=E_f

so:

mgh = \frac{1}{2}MV_i^2

where g is the gravity and h the altitude. So, replacing values, we get:

(85kg)(9.8m/s^2)(36m)= \frac{1}{2}(85kg)V_i^2

solving for V_i:

V_i = 26.56m/s

Then, replacing in the initial equation:

I = MV_f-MV_i

I = (85kg)(0m/s)-(85kg)(26.56m/s)

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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