Chapter 4.2, Problem 24SSC

To determine

The conclusion about the elevator’s motion.

Answer to Problem 24SSC

Itcan be concluded that the lift is moving downward with acceleration $0.5\text{m}/{\text{s}}^{\text{2}}$ downward.

Explanation of Solution

Given info:

The mass suspended is $m=1\text{kg}$ .

The reading of scale is $9.3\text{N}$ .

Formula used:

Since the elevator spring scale reads a different value it means the lift is moving.

Force due to earth’s mass can be obtained by,

  $F_{\text{earth's}\text{mass}}=\text{suspended}\text{}\text{mass}\times g$

Acceleration of elevator can be obtained by,

  $\text{acceleration}=\frac{F_{scale}-F_{earth\text{'}smass}}{\text{suspended mass}}$

Calculation:

Substitute the given values in equation (1).

  $\begin{array}{c}{F}_{earth\text{'}smass}=1\text{}\text{kg}\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\\ =9.80\text{N}\end{array}$

Substitute the given values in equation (2).

  $\begin{array}{c}\text{acceleration}=\frac{F_{scale}-F_{earth\text{'}smass}}{\text{suspended mass}}\\ =\frac{9.3\text{N}-9.80\text{N}}{1\text{kg}}\\ =-0.5\text{m}/{\text{s}}^{\text{2}}\end{array}$

Since, the acceleration is negative it shows that the lift is accelerating downward.

Conclusion:

Therefore, it can be concluded that the lift is moving downward with acceleration $0.5\text{m}/{\text{s}}^{\text{2}}$ downward.