91Ó°ÊÓ

The given data can be listed below as follows,

Newton’s second law gives the relationship between the force and the acceleration. This law allows us to calculate the acceleration of a given mass and known forces of an object.

(a)

The mass of the student alone whose weight is$W_s=550N$ is expressed as,

$m_s=\frac{W_s}{g}$

Here g is the acceleration due to gravity

Substitute$9.80m/s^2$ for g and 550 N for$W_s$ in the above equation, and we get,

$$\begin{gathered} m_s=\frac{550N}{9.80m/s^2} \\ {m}_s=56.1kg \end{gathered}$$

The relation of acceleration in terms of Newton’s second law is expressed as,

$F_y=m{a}_y$

which gives,

$$\begin{gathered} n-W_s=m_s{a}_y \\ {a}_y=\frac{n-W_s}{m_s} \end{gathered}$$

Here$a_y$ is the acceleration of the elevator,$W_s$ is the actual weight, n is the normal force exerted by the scale,$m_s$ is the mass of the student alone.

Substitute 450 N for n , 550 N for$W_s$ , 56.1 kg for$m_s$ in the above equation, and we get,

$$\begin{gathered} a_y=\frac{450N-550N}{56.1kg} \\ =-1.78m/s^2 \end{gathered}$$

Hence, the elevator has an acceleration of$1.78m/s^2$ , which is directed downward.

The relation of acceleration is expressed as,

$a_y=\frac{n-W_s}{m_s}$

Substitute 670 N for n , 550 N for $W_s$, and 56.1 kg for $m_s$in the above equation, and we get,

$$\begin{gathered} a_y=\frac{670N-550N}{56.1kg} \\ =2.14m/s^2 \end{gathered}$$

Hence, the acceleration is 2.14 $m/s^2$which is in an upward direction.

The relation of acceleration is expressed as,

$a_y=\frac{n-W_s}{m_s}$

Substitute n = 0 because scale reading is zero, 550 N for $W_s$, and 56.1 kg for $m_s$in the above equation, and we get,

$$\begin{gathered} a_y=\frac{0-550N}{56.1kg} \\ =-9.80m/s^2 \\ =-g \end{gathered}$$

Hence, the value of acceleration is equal to the negative of acceleration due to gravity which means the elevator is in free fall, and it is a troubled student should worry about this.

Newton’s second law is expressed as follows,

$F=m{a}_y$

which gives,

$$\begin{gathered} T-Mg=M{a}_y \\ T=M\left(g+a_y\right) \end{gathered}$$

In part (a)

Substitute -1.78$m/s^2$ for $a_y$, 850 kg for M ,$9.8m/s^2$ for g in the above equation, and we get,

$$\begin{gathered} T=\left(850kg\right)×\left(-1.78m/s^2+9.80m/s^2\right) \\ =6817N \end{gathered}$$

Hence, the tension in part (a) is 6817 N .

In part (c)

Substitute$-9.8m/s^2$ for g in the above equation , and we get,

$$\begin{gathered} T=\left(850N\right)×\left(-9.80m/s^2+9.80m/s^2\right) \\ =0 \end{gathered}$$

Hence, the tension in part (c) is zero.