Question

An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.4 s. A passenger in the elevator is holding a 3.3 kg bundle at the end of a vertical cord. What is the tension in the cord as the elevator accelerates?

Answer 1

35.71 N

Step-by-step explanation:

The elevator starts from the rest means its initial velocity is zero.

Given that, the height achieved by the elevator in 1.4 s will be,
`S=1m`

Given that the mass of the bundle which is hold by passenger is,
`m=3.3 kg`

Now according to second equation of motion.

`S=ut+(1)/(2)at^(2)`

Here, S is the height, u is the initial velocity, t is the time taken, and a is the acceleration.

Now initial velocity is zero therefore,

`S=(1)/(2)at^(2)\\a=(2S)/(t^(2) )`

According to the free body diagram tension and acceleration in upward direction and weight is in downward direction.

So,

`ma=T-mg\\T=m(g+a)`

Put the value of a from the above

`T=m(g+(2S)/(t^(2) ))`

Put all the variables.

`T=3.3(9.8+(2* 1)/(1.4^(2) ))\\T=3.3(9.8+1.02)\\t=35.71N`

This the required tension.