Question

It takes the elevator in a skyscraper 3.6 s to reach its cruising speed of 11 m/s. A 62 kg passenger gets aboard on the ground floor.

What is the passenger's weight before theelevator starts moving?

Answer 1

Weight of the passenger will be 607.6 N

Step-by-step explanation:

We have given that a elevator takes 3.6 sec to reach its cursing speed

So time t = 3.6 sec

Velocity v = 11 m /sec

Mass of the passenger m = 62 kg

We have to find the weight of the passenger weight before elevator starts moving

Before the elevator starts moving

`g_(net)=g=9.8m/sec^2`

So weight of the passenger will be equal to
`W=mg=62* 9.8=607.6N`