Question

A customer sits in an amusement park ride in which the compartment is to be pulled downward in the negative direction of a yaxis with an acceleration magnitude of$\mathbf{1}\mathbf{.}\mathbf{24}\mathbf{g}$, with$\mathit{g}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{80}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$. A $\mathbf{0}\mathbf{.}\mathbf{567}\mathbf{}\mathbf{g}$coin rests on the customer’s knee. Once the motion begins and in unit-vector notation, (a) what is the coin’s acceleration relative to the ground and (b) what is the coin’s acceleration relative to the customer? (c) How long does the coin take to reach the compartment ceiling,$\mathbf{2}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{m}$above the knee? In unit-vector notation, (d) what is the actual force on the coin and (e) what is the apparent force according to the customer’s measure of the coin’s acceleration?

Short answer

1. Acceleration of coin relative to the ground is$-9.8\mathrm{m}/{\mathrm{s}}^2$.
2. Acceleration of coin relative to the customer is$2.35\mathrm{m}/{\mathrm{s}}^2$.
3. Time taken by coin to reach compartment ceiling is$1.37\mathrm{s}$.
4. Actual forces on the coin is$-5.56\times {10}^{-3}\mathrm{N}$.
5. Apparent forces according to the costumer’s measure of the coin’s acceleration is$1.33\times {10}^{-3}\mathrm{N}$ .

Step-by-step solution

Given data

• Mass of coin is$m=0.567\mathrm{g}$ .
• The acceleration of customer while going downward is$a=1.24g$ , as$g=9.8\mathrm{m}/{\mathrm{s}}^2$
• Height of the celling is $2.20\mathrm{m}$above the knee.

Understanding the concept of the net force

Newton’s second law states that the net force $\stackrel{\mathbf{\rightharpoonup }}{\mathit{F}}$on a body with mass mis related to the body’s acceleration $\overrightarrow{\mathit{a}}$by,

$\stackrel{\mathbf{\rightharpoonup }}{\mathit{F}\mathbf{}}\mathbf{=}\mathit{m}\stackrel{\mathbf{\rightharpoonup }}{\mathit{a}}$

The customer sits on the amusement park ride which goes downward with an acceleration of 1.24g, and we are given the coin’s mass. By using the force equation and kinematic equation, we can find actual and apparent acceleration on the coin. By using the kinematic equation, we can find the time taken by the coin to reach a certain height. Using the value of acceleration, we can find the value of force.

Formulae:

$F=ma$ (i)

$s=v_{°}t+\frac{1}{2}a{t}^2$ (ii)

(a) Calculate the acceleration of the coin while going downward.

When the object is moving downward, the free fall is

$$\begin{gathered} a_{coin}=-g \\ =-9.8\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration of the coin is$-9.8\mathrm{m}/{\mathrm{s}}^2$ .

(b) Calculate the acceleration of the coin relative to the customer

The customer is pulled down with acceleration,

$$\begin{gathered} 1.24g=1.24\times 9.8\mathrm{m}/{\mathrm{s}}^2 \\ =12.15\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

So, the real acceleration if the coin is

$$\begin{gathered} a_{rel}=a_{coin}-a_{customer} \\ =(-9.8\mathrm{m}/{\mathrm{s}}^2)-(-12.15\mathrm{m}/{\mathrm{s}}^2) \\ =2.35\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The acceleration is positive and it means it is moving in the positive y-direction

(c) Calculate the time taken by coin to reach the ceiling

The height of the ceiling above the knee is2.20 m . Therefore, fromthe equation (ii),

$$\begin{gathered} s=v_{°}t+\frac{1}{2}{a}_{rel}{t}^2 \\ 2.20m=0+\frac{1}{2}(2.35\mathrm{m}/{\mathrm{s}}^2){\mathrm{t}}^2 \\ \mathrm{t}=\sqrt{\frac{(2.20)\left(2\right)}{2.35\mathrm{m}/{\mathrm{s}}^2}} \\ =1.37\mathrm{s} \end{gathered}$$

Therefore, the coin takes 1.37 to reach the ceiling.

(d) Calculate the actual force on the coin

The actual force is the product of mass and the actual acceleration on the coin. From the equation (i)

$$\begin{gathered} F_{coin}\left(\mathrm{Actual}\right)=(0.567\times {10}^{-3}\mathrm{kg})(-9.8\mathrm{m}/{\mathrm{s}}^2) \\ =-5.56\times {10}^{-3}\mathrm{N} \end{gathered}$$

Therefore, the actual force on the coin is$-5.56\times {10}^{-3}\mathrm{N}$ .

(e) Calculate the apparent force on coin

In the customer’s frame of reference, the coin is moving upward direction so, the apparent force on the coin,

$$\begin{gathered} F_{coin}\left(\mathrm{Apparent}\right)=(0.567\times {10}^{-3}\mathrm{kg})(2.35\mathrm{m}/{\mathrm{s}}^2) \\ =1.33\times {10}^{-3}\mathrm{N} \end{gathered}$$

Therefore, apparent force on the coin is $1.33\times {10}^{-3}\mathrm{N}$.