What speed should a roller coaster have at a point 30 m above the bottom of a dip if it is to barely reach the top of the next slope, which is 40 m above the bottom of the dip? Neglect friction. A) 0 m/s B) 14 m/s C) 28 m/s D) 98 m/s E) 2.0* 10² m/s

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To solve this, we use the principle of conservation of mechanical energy, which states that the total mechanical energy (potential energy + kinetic energy) in a system remains constant if only conservative forces are acting. At the top of the next slope, the roller coaster needs just enough kinetic energy to reach but not exceed the height of the slope, implying it would have a speed of 0 m/s at the top. Thus, all its energy at the point 30 m above the bottom would be converted into potential energy at the 40 m height.The potential energy (PE) at a height $h$ is given by $PE = mgh$, where $m$ is the mass, $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$), and $h$ is the height above the reference point. The kinetic energy (KE) is given by $KE = \frac{1}{2}mv^2$, where $v$ is the velocity.At the 30 m point, the roller coaster has potential energy due to its height and needs kinetic energy to reach the 40 m height. The energy conservation between the two points (30 m and 40 m) can be expressed as:$$mgh_1 + \frac{1}{2}mv^2 = mgh_2$$Given $h_1 = 30 \, \text{m}$, $h_2 = 40 \, \text{m}$, and $g = 9.8 \, \text{m/s}^2$, we find $v$. The mass $m$ cancels out from both sides, simplifying to:$$9.8 \times 30 + \frac{1}{2}v^2 = 9.8 \times 40$$Solving for $v$, we find $v = 14 \, \text{m/s}$. Therefore, the roller coaster must have a speed of 14 m/s at the 30 m point to just reach the top of the next slope at 40 m.

What Speed Should a Roller Coaster Have at a Point