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Q7.3-7 CQ

Expert-verifiedFound in: Page 258

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**In Example 7.7, we calculated the final speed of a roller coaster that descended \(20{\rm{ m}}\) in height and had an initial speed of \(5{\rm{ m}}/{\rm{s}}\) downhill. Suppose the roller coaster had had an initial speed of \(5{\rm{ m}}/{\rm{s}}\) uphill instead, and it coasted uphill, stopped, and then rolled back down to a final point 20m below the start. We would find in that case that it had the same final speed. Explain in terms of conservation of energy.**

The kinetic energy of the coaster and the potential energy of the coaster all remains constant. Therefore, the final velocity of the coaster will remain the same whether the coaster moves up or down the hill.

**Conservation of energy: According to the conservation of energy, energy neither be created nor be destroyed. The system's total energy is always conserved.**

When the coaster starts to descend, its potential energy gets converted into kinetic energy. Mathematically,

\(\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = mgh\)

Here, \(m\) is the mass of the coaster, \(g\) is the acceleration due to gravity \(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\), \({v_f}\) is the final velocity of the coaster, \({v_i}\) is the initial velocity of the coaster \(\left( {{v_i} = 5{\rm{ m}}/{\rm{s}}} \right)\), and h is the height \(\left( {h = 20.0{\rm{ m}}} \right)\).

As a result, the final velocity of the coaster is given as,

\({v_f} = \sqrt {v_i^2 + 2gh} \)

Putting all known values,

\(\begin{aligned}{c}{v_f} = \sqrt {{{\left( {5{\rm{ m}}/{\rm{s}}} \right)}^2} + 2 \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {20.0{\rm{ m}}} \right)} \\ = 20.42{\rm{ m}}/{\rm{s}}\end{aligned}\)

When the coaster ascends uphill, its kinetic energy gets converted into potential energy.

\(\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = mgh'\)

Here, \(m\) is the mass of the coaster, \(g\) is the acceleration due to gravity \(\left( {g = - 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\), \({v_f}\) is the final velocity of the coaster \(\left( {{v_f} = 0} \right)\), \({v_i}\) is the initial velocity of the coaster \(\left( {{v_i} = 5{\rm{ m}}/{\rm{s}}} \right)\), and \(h'\) is the height above the starting point.

Therefore, the height above the starting point attained by the coaster is,

\(h' = \frac{{v_f^2 - v_i^2}}{{2g}}\)

Putting all known values,

\(\begin{aligned}{}h' = \frac{{{{\left( 0 \right)}^2} - {{\left( {5{\rm{ m}}/{\rm{s}}} \right)}^2}}}{{2 \times \left( { - 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}\\ = 1.28{\rm{ m}}\end{aligned}\)

When the coaster stops and starts to roll down, its initial velocity becomes zero \(\left( {{v_i} = 0} \right)\), and it is a height of,

\(\begin{aligned}{}H = h + h'\\ = \left( {20.0{\rm{ m}}} \right) + \left( {{\rm{1}}{\rm{.28 m}}} \right)\\ = 21.28{\rm{ m}}\end{aligned}\)

While rolling down the potential energy of the coaster gets converted into kinetic energy. Mathematically,

\(\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = mgH\)

As a result, the final velocity of the coaster is given as,

\({v_f} = \sqrt {v_i^2 + 2gH} \)

Putting all known values,

\(\begin{aligned}{}{v_f} = \sqrt {{{\left( 0 \right)}^2} + 2 \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {21.28{\rm{ m}}} \right)} \\ = 20.42{\rm{ m}}/{\rm{s}}\end{aligned}\)

Therefore, in both cases the coaster will descend down with same speed.

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