It takes the elevator in a skyscraper 4.1 S to reach its cruising speed of 12 m/

Question

It takes the elevator in a skyscraper 4.1 S to reach its cruising speed of 12 m/s . A 70kg passenger gets aboard on the ground floor. Part A What is the passengers weight before the elevator starts moving? Express your answer to two significant figures and include the appropriate units. ? Part B What is the passenger?s weight while the elevator is speeding up? Express your answer to two significant figures and include the appropriate units. ? Part C What is the passenger?s weight after the elevator reaches its cruising speed? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

time=4.1 sec

mass=70 kg

speed=12 m/s

We know that the weight of the passengers is:

weight = 70 * g

= 70* 9,81

= 686.7 N

2)Assuming that the lift accelerations at a constant rate, the acceleration of the lift is:

a = (v - u) / t = (12- 0) / 4.1 = 12 / 4.1 = 2.92 m/s2

The upward force from the lift on the passenger is

F = ma

= 70* 2.92

= 204.87 N

This means that the effective weight of the passenger is

w = 686.7 + 204.87

= 891.57 N

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