Given,
The weight of the student, W=500 N
Thus the mass of the student is given by,
[tex]m=\frac{W}{g}[/tex]Where g is the acceleration due to gravity,
On substituting the known values,
[tex]\begin{gathered} m=\frac{500}{9.8} \\ =51.02\text{ kg} \end{gathered}[/tex]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,
[tex]F=ma[/tex]Where a is the acceleration of the elevator.
On substituting the known values,
[tex]\begin{gathered} 200=51.02\times a \\ \Rightarrow a=\frac{200}{51.02} \\ =3.92m/s^2 \end{gathered}[/tex]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,
[tex]\begin{gathered} F=ma \\ \Rightarrow a=\frac{F}{m} \\ a=\frac{-200}{51.02} \\ =-3.92m/s^2 \end{gathered}[/tex]Thus the accelerating in this interval is -3.92 m/s²That is the elevator is accelerating downwards.
