Question

Calculate the acceleration of the elevator for each 5 second interval

Answer 1

Given,

The weight of the student, W=500 N

Thus the mass of the student is given by,

`m=(W)/(g)`

Where g is the acceleration due to gravity,

On substituting the known values,

`\begin{gathered} m=(500)/(9.8) \\ =51.02\text{ kg} \end{gathered}`

For the first 5 seconds, the net force acting on the student is 0 N. Thus scale reads only his weight. As the net force is zero the acceleration of the elevator is also zero.

In the next 5 intervals, the net force acting on the student is F=200 N, as seen from the diagram.

Thus the acceleration of the elevator is given by the equation,

`F=ma`

Where a is the acceleration of the elevator.

On substituting the known values,

`\begin{gathered} 200=51.02* a \\ \Rightarrow a=(200)/(51.02) \\ =3.92m/s^2 \end{gathered}`

Thus the acceleration in this interval is 3.92 m/s²

During the interval, 10s-15s, the net force acting on the student is zero as seen from the graph. Thus the acceleration of the elevator is also zero.

During the interval, 15 s-20 s, the net force on the student is F=-200 N as seen from the diagram.

Thus the acceleration is,

`\begin{gathered} F=ma \\ \Rightarrow a=(F)/(m) \\ a=(-200)/(51.02) \\ =-3.92m/s^2 \end{gathered}`

Thus the accelerating in this interval is -3.92 m/s²That is the elevator is accelerating downwards.