Question

While an elevator of mass 2325 kg moves upward, the tension in the cable is 38.8 kN. Assume the elevator is supported by a single cable. Forces exerted by the guide rails and air resistance are negligible.the acceleration of the elevator in the upward direction = 6.88 m/s2If at some point in the motion the velocity of the elevator is 1.20 m/s upward, what is the upward velocity of the elevator 4.00 s later?

Answer 1

Given data:

* The acceleration of the elevator is,

`a=6.88ms^(-2)`

* The initial velocity of the elevator is u = 1.2 m/s.

* The time taken by the elevator is t = 4 seconds.

Solution:

By the kinematics equation, the final velocity of the elevator is,

`v-u=at`

where v is the final velocity,

`\begin{gathered} v-1.2=6.88*4 \\ v-1.2=27.52 \\ v=27.52+1.2 \\ v=28.72\text{ m/s} \end{gathered}`

Thus, the final velocity of the elevator in the upward direction is 28.72 m/s in the upward direction.