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

Answer:

35.71 N

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.

$S=ut+\frac{1}{2}at^{2}$

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,

$S=\frac{1}{2}at^{2}\\a=\frac{2S}{t^{2} }$

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

$T=m(g+\frac{2S}{t^{2} })$

Put all the variables.

$T=3.3(9.8+\frac{2\times 1}{1.4^{2} })\\T=3.3(9.8+1.02)\\t=35.71N$

This the required tension.