Question

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

Answer 1

During the first 5 seconds, the velocity is 0 m/s

During the next 5 s (between 5 s and 10 s), the velocity is 19.6 m/s

During the next 5 s (between 10 s and 15 s), the velocity is 0 m/s.

During the next 5 s (between 15 s and 20 s), the velocity is -19.6 m/s.

How to calculate the velocity of the student?

The velocity of the student at the end of each 5 second interval is calculated by applying the following formula.

F(net) = ma

F(net) = mv/t

v = F(net) x t / m

The mass, m is calculated as;

W = mg

m = W / g

m = (500 N ) / 9.8 m/s²

m = 51.02 kg

The net force during the first 5 seconds is 0 N, and the velocity is calculated as;

v = (0 x 5 ) / 51.02

v = 0 m/s

During the next 5 s (between 5 s and 10 s),

F(net) = 700 N - 500 N = 200 N

v = (200 x 5 ) / 51.02

v = 19.6 m/s

During the next 5 s (between 10 s and 15 s),

F(net) = 500 N - 500 N = 0 N

v = (0 x 5 ) / 51.02

v = 0 m/s

During the next 5 s (between 15 s and 20 s),

F(net) = 300 N - 500 N = -200 N

v = (-200 x 5 ) / 51.02

v = -19.6 m/s

Answer 2

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=(W)/(g)`

Then, the acceleration at every moment is given by:

`a=(F_N-W)/(m)=(F_N-W)/(((W)/(g)))=((F_N-W)/(W))* g`

In the first interval, the acceleration is:

`a_1=(500N-500N)/(500N)*9.8(m)/(s^2)=0`

In the second interval, the acceleration is:

`\begin{gathered} a_2=(700N-500N)/(500N)*9.8(m)/(s^2) \\ =3.92(m)/(s^2) \end{gathered}`

In the third interval, the acceleration is:

`a_3=(500N-500N)/(500N)*9.8(m)/(s^2)=0`

And in the fourth interval, the acceleration is:

`a_4=(300N-500N)/(500N)*9.8(m)/(s^2)=-3.92(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*5s=0 \\ v_2=v_1+a_2t=0+3.92(m)/(s^2)*5s=19.6(m)/(s) \\ v_3=v_2+a_3t=19.6(m)/(s)+0*5s=19.6(m)/(s) \\ v_4=v_3+a_4t=19.6(m)/(s)-3.92(m)/(s^2)*5s=0 \end{gathered}`

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

`\begin{gathered} v_(5s)=0(m)/(s) \\ v_(10s)=19.6(m)/(s) \\ v_(15s)=19.6(m)/(s) \\ v_(20s)=0(m)/(s) \end{gathered}`