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?

• 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.

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)

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$-9.8\mathrm{m}/{\mathrm{s}}^2$ .

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

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.

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$-5.56\times {10}^{-3}\mathrm{N}$ .

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 $1.33\times {10}^{-3}\mathrm{N}$.