Chapter 7, Problem 61P

To determine

To Find:The apparent weight of a $450N$ passenger at points B and C.

Answer to Problem 61P

The apparent weight at point B is $3900N$ and apparent weight at point C is $2100N$ .

Explanation of Solution

Given:

The roller coaster has initial speed of $7.00m/s$ at point A and friction is neglected.

Formula Used:

Kinetic energy is given by formula

  $KE=\frac{1}{2}m{v}^2$

Where,

  $m=$ Mass

  $v=$ Velocity

Potential energy is given by formula

  $PE=mgh$

Where,

  $m=$ Mass

  $g=$ Acceleration due to gravity

  $h=$ Height

Calculation:

Using Law of conservation of energy at point B

  $mg{h}_A=\frac{1}{2}m{v}_B{}^2-\frac{1}{2}m{v}_A{}^2$

On solving the above equation

  $\begin{array}{l}g{h}_A=\frac{1}{2}{v}_B{}^2-\frac{1}{2}{v}_A{}^2\\ \frac{1}{2}{v}_B{}^2=g{h}_A+\frac{1}{2}{v}_A{}^2\\ v_B{}^2=2g{h}_A+v_A{}^2\\ v_B=\sqrt{2g{h}_A+v_A{}^2}-----(1)\end{array}$

Here,

  $\begin{array}{l}{v}_A=7.00m/s\\ h_A=9.00m\\ g=9.8m/s^2\end{array}$

Substitute the values in Equation (1),

  $\begin{array}{l}{v}_B=\sqrt{2\left(9.8\right)\left(9\right)+{\left(7\right)}^2}\\ v_B=\sqrt{176.4+49}\\ v_B=\sqrt{225.4}\\ v_B=15.01m/s\end{array}$

Now, apparent weight at point B is calculated as

  $W_B=W\left(1+\frac{v_B{}^2}{rg}\right)$

Substitute the values

  $\begin{array}{l}{W}_B=450\left(1+\frac{{\left(15.01\right)}^2}{3\times 9.8}\right)\\ W_B=450\left(1+7.67\right)\\ W_B=450\left(8.67\right)\\ W_B\approx 3900N\end{array}$

Using Law of conservation of energy at point C

  $mg{h}_A=mg{h}_B+\frac{1}{2}m{v}_C{}^2-\frac{1}{2}m{v}_A{}^2$

On solving the above equation

  $\begin{array}{l}g{h}_A=g{h}_B+\frac{1}{2}{v}_C{}^2-\frac{1}{2}{v}_A{}^2\\ \frac{1}{2}{v}_C{}^2=g{h}_A-g{h}_B+\frac{1}{2}{v}_A{}^2\\ v_C{}^2=v_A{}^2+2g\left(h_A-h_B\right)\\ v_C=\sqrt{v_A{}^2+2g\left(h_A-h_B\right)}-----(2)\end{array}$

Here,

  $\begin{array}{l}{v}_A=7.00m/s\\ h_A=9.00m\\ h_B=6.00m\\ g=9.8m/s^2\end{array}$

Substituting the values in Equation (2),

  $\begin{array}{l}{v}_C=\sqrt{{\left(7\right)}^2+2\left(9.8\right)\left(9-6\right)}\\ v_C=\sqrt{49+2\left(9.8\right)\left(3\right)}\\ v_C=\sqrt{49+58.8}\\ v_C=\sqrt{107.8}\\ v_C=10.38m/s\end{array}$

Now, apparent weight at point C is calculated as

  $W_C=W\left(1+\frac{v_C{}^2}{rg}\right)$

Substituting the values,

  $\begin{array}{l}{W}_C=450\left(1+\frac{{\left(10.38\right)}^2}{3\times 9.8}\right)\\ W_C=450\left(1+3.67\right)\\ W_C=450\left(4.67\right)\\ W_C\approx 2100N\end{array}$

Conclusion:

Thus, apparent weight at point B is $3900N$ and apparent weight at point C is $2100N$ .