Chapter 6, Problem 50P

To determine

The speed which theroller coaster reaches point 2.

Answer to Problem 50P

Solution:

Speed of roller coaster at point 2 $\text{=20}\text{.6}\text{m/s}$

Explanation of Solution

Given:

Height of point $1$ from reference $\text{=}\text{32}\text{m}$

Height of point $2$ , from reference $\text{= 0}\text{m}$

Point 2 is taken as reference point

Speed of ball at point

  $\text{1 }\left({\text{v}}_{\text{1}}\right)\text{=1}\text{.30}\text{m/s}$

Distance travelled by roller $\text{-coaster}\text{=}\text{45}\text{.0}\text{m}$

Friction force $\text{=0}\text{.23}\left(W_r\right)$

  $W_r$: Weight of roller coaster

FormulaUsed:

  $W_{NC}=\Delta KE+\Delta PE$

  $W_{NC}\text{:}$ Work done by non-conservative forces.

  $\Delta$ K.E : change in kinetic energy

  $\Delta$ P.E : change in potential energy

  $W_f=-F_f\times S$

  $F_f:$ Frictional force

S : distance travelled

Calculation:

Potential energy at point 1 $\text{= }mg{h}_1$

  $P.E_1=m\times 9.8\times 32$

Potential energy at point 2 $=mg{h}_2$

  $\begin{array}{l}P{E}_2=m\times \text{9}\text{.8×0}\\ \text{ =0}\end{array}$

Kinetic energy at point 1 $=\frac{1}{2}m{v}_1{}^2$

  $=\frac{1}{2}\times m\times {\left(1.30\right)}^2$

Kinetic energy at point 2 $=\frac{1}{2}m{v}_2{}^2$

  $=\frac{1}{2}\times m{v}_2{}^2$

  $\begin{array}{l}{W}_{NC}=-0.23\left(W_r\right)\left(45.0\right)\\ =-0.23mg\times 45.0\mathrm{...}(2)\\ \frac{1}{2}m{v}_2{}^2-\frac{1}{2}m{\left(1.30\right)}^2-m\times 9.8\times 32=-0.23mg\times 45.0\\ \frac{1}{2}{v}_2{}^2-\frac{1}{2}{\left(1.30\right)}^2-9.8\times 32=-0.23\times 9.8\times 45\\ v_2=\sqrt{2\times 9.8\times 32-0.23\times 9.8\times 45\times 2+{\left(1.30\right)}^2}\\ =20.64\text{m/s}\\ =20.6\text{m/s}\end{array}$

Calculation:

Work done by friction force depends upon distance travelled not displacement. Here work done by friction force is negative.Itconverts mechanical energy into thermal energy.