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Chapter 7: Problem 25

A particle slides back and forth on a frictionless track whose height as a function of horizontal position \(x\) is \(y=a x^{2},\) where \(a=0.92 \mathrm{m}^{-1} .\) If the particle's maximum speed is \(8.5 \mathrm{m} / \mathrm{s},\) find its turning points.

Short Answer

Expert verified
The turning points are at \(x = \pm\sqrt{36.125 J / (9.8 m/s^{2} * 0.92 m^{-1})}\) m.

Step by step solution

01

Calculate Kinetic Energy

Initially, the whole energy is kinetic energy. The formula for kinetic energy is \(KE = (1/2) m v^{2}\). Since the mass of the particle doesn't affect this problem, we are able to omit it. Therefore, our initial kinetic energy will be \(KE = (1/2) * (8.5 m/s)^{2} = 36.125 J\).
02

Translate kinetic energy to potential energy

The potential energy at the turning points (velocity=0) is equal to the initial kinetic energy. Potential energy can be calculated by \(PE = m * g * h\), and in our case the height \(h = y = a * x^{2}\). As stated earlier, the mass is not affecting this problem and can be omitted. Thus, we can solve for x the equation \(PE = g * a * x^{2} = 36.125 J\).
03

Solve for x

To solve this equation, first remember that g (the acceleration due to gravity) is approximately 9.8 m/s^{2}. Then solve the equation \(9.8 m/s^{2} * 0.92 m^{-1} * x^{2} = 36.125 J\), resulting in \(x^{2} = 36.125 J / (9.8 m/s^{2} * 0.92 m^{-1})\), and thus \(x = \pm\sqrt{36.125 J / (9.8 m/s^{2} * 0.92 m^{-1})}\) since x can be positive or negative as the particle can go to the left or to the right.

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Most popular questions from this chapter

A particle of mass \(m\) is subject to a force \(\vec{F}=(a \sqrt{x}) \hat{\imath},\) where \(a\) is a constant. The particle is initially at rest at the origin and is given a slight nudge in the positive \(x\) -direction. Find an expression for its speed as a function of position \(x\)

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