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Chapter 5: Problem 32

A \(50-\mathrm{kg}\) pole vaulter running at \(10 \mathrm{~m} / \mathrm{s}\) vaults over the bar. Her speed when she is above the bar is \(1.0 \mathrm{~m} / \mathrm{s}\). Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.

Short Answer

Expert verified
The pole vaulter crosses the bar at an approximate altitude of 51.02 meters.

Step by step solution

01

Identify the Principle

The concept used in this exercise is the conservation of energy. As the pole vaulter runs and vaults over the bar, kinetic energy (due to her movement) converts into potential energy (due to her height / altitude).
02

Formulate the Equation

By using the principle of conservation of energy, the initial kinetic energy should equal the final potential energy. The initial kinetic energy (KE) can be calculated by using the formula \(\frac{1}{2} m v^{2}\), where m is mass and v is velocity. The final potential energy (PE) can be calculated by using the formula \(m g h\), where m is mass, g is acceleration due to gravity and h is height.
03

Initial Kinetic Energy

Calculate the initial kinetic energy using the formula. The mass 'm' of the vaulter is 50 kg and her initial velocity 'v' is 10 m/s. KE = \(\frac{1}{2} m v^{2} = \frac{1}{2} \times 50 kg \times (10 m/s)^{2} = 2500 J (joules).
04

Final Potential Energy

The final potential energy should equal the initial kinetic energy, as energy is conserved. The speed of the vaulter when she crosses the bar is given as 1.0 m/s. However, this speed does not contribute to her altitude and can be neglected in calculating potential energy, as the vertical component is zero at maximum height. So, PE (Final) = KE (Initial) = 2500 J.
05

Height Calculation

Now, we will rearrange the equation for potential energy to solve for height. \(h = \frac{PE}{mg}\), where 'g' is the acceleration due to gravity. Using the standard value of g = 9.8 m/(s)^2, we get, h = \(\frac{2500 J}{50 kg \times 9.8 m/(s)^2} = 51.02 m\).

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Most popular questions from this chapter

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