Question

Determine the velocity V of the elevator at the end of each 5-second interval.

Answer 1

The reading of the scale corresponds to the normal force that the floor of the elevator exerts over the student.

The net force is always equal to the mass of the student times its acceleration, according to Newton's Second Law of Motion:

$F_N-W=ma$

When the elevator is at rest, the acceleration is 0 and the reading of the scale is equal to the weight of the student.

On the other hand, the mass of the student is given by:

$m=\frac{W}{g}$

Then, the acceleration at every moment is given by:

$a=\frac{F_N-W}{m}=\frac{F_N-W}{(\frac{W}{g})}=(\frac{F_N-W}{W})\times g$

In the first interval, the acceleration is:

$a_1=\frac{500N-500N}{500N}\times9.8\frac{m}{s^2}=0$

In the second interval, the acceleration is:

$\begin{gathered} a_2=\frac{700N-500N}{500N}\times9.8\frac{m}{s^2} \\ =3.92\frac{m}{s^2} \end{gathered}$

In the third interval, the acceleration is:

$a_3=\frac{500N-500N}{500N}\times9.8\frac{m}{s^2}=0$

And in the fourth interval, the acceleration is:

$a_4=\frac{300N-500N}{500N}\times9.8\frac{m}{s^2}=-3.92\frac{m}{s^2}$

We can find the velocity at the end of each interval using the formula:

$v_f=v_0+at$

For t=5s, v_0=0 and the corresponding accelerations, we have:

$\begin{gathered} v_1=a_1t=0\times5s=0 \\ v_2=v_1+a_2t=0+3.92\frac{m}{s^2}\times5s=19.6\frac{m}{s} \\ v_3=v_2+a_3t=19.6\frac{m}{s}+0\times5s=19.6\frac{m}{s} \\ v_4=v_3+a_4t=19.6\frac{m}{s}-3.92\frac{m}{s^2}\times5s=0 \end{gathered}$

Therefore, the velocities after each 5-seconds interval, are:

$\begin{gathered} v_{5s}=0\frac{m}{s} \\ v_{10s}=19.6\frac{m}{s} \\ v_{15s}=19.6\frac{m}{s} \\ v_{20s}=0\frac{m}{s} \end{gathered}$