Q45P - University Physics with Modern Physics

Question

After an annual checkup, you leave your physician’s office, where you weighed 683 N . You then get into an elevator that, conveniently, has a scale. Find the magnitude and direction of the elevator’s acceleration if the scale reads (a) 725 N and (b) 595 N.

Step-by-Step Solution

Answer

(a) The magnitude of the elevator’s acceleration is $0.602\mathrm{m}/{\mathrm{s}}^2$ and it is in the upwards direction.

(b) The magnitude of the elevator’s acceleration is $-1.26\mathrm{m}/{\mathrm{s}}^2$ and it is in the downwards direction as indicated by negative sign.

Step 1: Identification of the given data

The given data is listed below as:

 

• The weight of yours is, w = 683 N
• The first reading of the scale is, $n_1=725 N$
• The second reading of the scale is, ${\mathrm{n}}_2=595 \mathrm{N}$ 

Step 2: Significance of the Newton’s second law

Newton’s second law states that the force applied to an object is directly proportional to the mass and the acceleration of the object. The weight of that object is also proportional to the mass of that object.

Step 3: (a) Determination of the magnitude and direction of the elevator’s acceleration in the first case

The equation of the acceleration of the elevator is expressed as:

$$\begin{gathered} n_1-w=ma \\ a=\frac{n_1-w}{m} \end{gathered}$$ 

 

Here, a is the acceleration of the elevator in the first case,$n_1$is the first reading of the first scale, w is your weig ht and m is your mass.

 

Substitute$\frac{w}{g}$ for m in the above equation.

 

$$\begin{gathered} n_1-w=ma \\ a=\frac{n_1-w}{{\displaystyle \frac{w}{g}}} \end{gathered}$$                                                                                                              …(i)

 

Here, g is the acceleration due to gravity with value $9.8 m/s^2$.

 

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{a}=\frac{725\mathrm{N}-683 \mathrm{N}}{{\displaystyle \frac{683 \mathrm{N}}{9.8 \mathrm{m}/{\mathrm{s}}^2}}} \\ =\frac{42 \mathrm{N}}{683 \mathrm{N}}\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right) \\ =0.06\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right) \\ =0.602 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

As the acceleration is positive, hence it is directed upwards.

 

Thus, the magnitude of the elevator’s acceleration is$0.602 \mathrm{m}/{\mathrm{s}}^2$and it is in the upwards direction.

Step 4: (b) Determination of the magnitude and direction of the elevator’s acceleration in the second case

Substitute all the values in equation (i).

$$\begin{gathered} \mathrm{a}=\frac{595 \mathrm{N}-683 \mathrm{N}}{{\displaystyle \frac{683 \mathrm{N}}{9.8 \mathrm{m}/{\mathrm{s}}^2}}} \\ =\frac{-88 \mathrm{N}}{683 \mathrm{N}}\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right) \\ =-0.12\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right) \\ =-1.26 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

As the acceleration is positive, hence it is directed downwards.

 

Thus, the magnitude of the elevator’s acceleration is $-1.26 \mathrm{m}/{\mathrm{s}}^2$ and it is in the downwards direction.