91影视

• 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鈥檚 second law states that the net force $\stackrel{\mathbf{鈬赌}}{\mathit{F}}$on a body with mass mis related to the body鈥檚 acceleration $\stackrel{\mathbf{鈫�}}{\mathit{a}}$by,

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

The customer sits on the amusement park ride which goes downward with an acceleration of 1.24g, and we are given the coin鈥檚 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)

$s=v_{掳}t+\frac{1}{2}a{t}^2$ (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,

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

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),

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

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)

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

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

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

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

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