Q8E - University Physics with Modern Physics

Question

You walk into an elevator, step onto a scale, and push the “up” button. You recall that your normal weight is 625 N. Draw a free-body diagram. 

 

(a) When the elevator has an upward acceleration of magnitude $2.5 m/S^2$, what does the scale read? 

 

(b) If you hold a 3.85 kg package by a light vertical string, what will be the tension in this string when the elevator accelerates as in part (a)?

Step-by-Step Solution

Answer

The free body diagram is as follows:

(a) When the elevator has an upward acceleration of magnitude $2.5 m/s^2$ , the scale reads  $80.05 kg$.

(b) If a 3.85 kg package is held by a light vertical string inside the accelerating elevator, the tension in this string will be $47.36 N$ .

 

 

Step 1: Given data

The normal weight of the person stepping in the elevator is

  $W=625 \text{N}$

The acceleration of the elevator is

  $a=2.5 {\text{m/s}}^2$

The mass of the package is

  $m=3.85 \text{kg}$

Step 2: Net weight inside an elevator

The weight of a body of mass   inside an elevator moving up with an acceleration a  is

 $W=m\left(g+a\right) .....\left(1\right)$   

Here,  g is the acceleration due to gravity.

Step 3: Weight of the person

The mass of the person is

  $\begin{array}{rcl}M& =& \frac{W}{g}\\ & =& \frac{625 \text{N}}{9.8 {\text{m/s}}^{\text{2}}}\\ & =& 63.78 \text{kg}\\ & & \end{array}$

From equation (1), the weight inside the elevator is

  $\begin{array}{rcl}W\text{'}& =& M\left(g+a\right)\\ & =& 63.78 \text{kg}\times \left(9.8 {\text{m/s}}^{\text{2}}+2.5 {\text{m/s}}^{\text{2}}\right)\\ & =& 784.5 \text{N}\\ & & \end{array}$

The scale will read

$\begin{array}{rcl}M\text{'}& =& \frac{W\text{'}}{g}\\ & =& \frac{784.5 \text{N}}{9.8 {\text{m/s}}^{\text{2}}}\\ & =& 80.05 \text{kg}\\ & & \end{array}$

The scale reads is 80.05 Kg

Step 3: Tension in the string

Tension in the string is equal to the weight of the object.

From equation (1), the weight of the object inside the elevator is

  $$\begin{gathered} W=m\left(g+a\right) \\ =3.85 \text{kg}\times \left(9.8 {\text{m/s}}^{\text{2}}+2.5 {\text{m/s}}^{\text{2}}\right) \\ =47.36 \text{N} \end{gathered}$$

Thus, the tension is 47.36N