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Chapter 6: Problem 41

Two pole-vaulters just clear the bar at the same height. The first lands at a speed of \(8.90 \mathrm{~m} / \mathrm{s},\) and the second lands at a speed of \(9.00 \mathrm{~m} / \mathrm{s} .\) The first vaulter clears the bar at a speed of \(1.00 \mathrm{~m} / \mathrm{s}\). Ignore air resistance and friction and determine the speed at which the second vaulter clears the bar.

Short Answer

Expert verified
The second pole-vaulter clears the bar at a speed of 1.67 m/s.

Step by step solution

01

Understanding the Problem

We are given two pole-vaulters who clear the bar at the same height. The first lands with a speed of 8.90 m/s and clears the bar at 1.00 m/s. The second lands with a speed of 9.00 m/s. We need to find the second vaulter's speed as they clear the bar.
02

Applying Conservation of Energy

For both vaulters, potential energy at the top, plus kinetic energy at the top, equals kinetic energy at landing. This is because we assume no energy loss due to air resistance or friction.
03

Setting Up the Equation for the First Vaulter

The energy conservation equation for the first vaulter is: \[ \frac{1}{2} m v_{1i}^2 + mgh = \frac{1}{2} m v_{1f}^2 \] where \(v_{1i} = 1.00 \text{ m/s}\) is the speed at the top, \(v_{1f} = 8.90 \text{ m/s}\) is the landing speed, \(h\) is the height of the bar, \(m\) is the mass, and \(g\) is the acceleration due to gravity.
04

Solving for the Height of the Bar

We solve for \(h\) by rearranging the first vaulter's equation: \[ mgh = \frac{1}{2} m v_{1f}^2 - \frac{1}{2} m v_{1i}^2 \] which simplifies to \[ gh = \frac{1}{2} (v_{1f}^2 - v_{1i}^2) \]Then, substitute the values: \(v_{1f} = 8.90 \text{ m/s}\) and \(v_{1i} = 1.00 \text{ m/s}\):\[ gh = \frac{1}{2} (8.90^2 - 1.00^2) \]
05

Calculating the Height

Calculate \(h\): \[ h = \frac{1}{2g} (8.90^2 - 1.00^2) \] With \(g = 9.81 \text{ m/s}^2\), this becomes: \[ h = \frac{1}{2 \times 9.81} (79.21 - 1.00) \] \[ h = \frac{1}{19.62} \times 78.21 \] \[ h = 3.99 \text{ m} \] (rounded to two decimal places).
06

Setting Up the Equation for the Second Vaulter

Write the energy conservation equation for the second vaulter:\[ \frac{1}{2} m v_{2i}^2 + mgh = \frac{1}{2} m v_{2f}^2 \] where \(v_{2f} = 9.00 \text{ m/s}\) is the landing speed and \(v_{2i}\) is the speed at the bar which we need to find.
07

Solving for the Speed at the Bar for the Second Vaulter

Rearrange to find \(v_{2i}\):\[ \frac{1}{2} m v_{2i}^2 = \frac{1}{2} m v_{2f}^2 - mgh \] \[ v_{2i}^2 = v_{2f}^2 - 2gh \] Substitute the values: \(v_{2f} = 9.00\) m/s and \(h = 3.99\) m:\[ v_{2i}^2 = 9.00^2 - 2 \times 9.81 \times 3.99 \] \[ v_{2i}^2 = 81.00 - 78.21 \] \[ v_{2i}^2 = 2.79 \] Taking the square root gives \(v_{2i} = \sqrt{2.79} = 1.67 \text{ m/s} \) (rounded).
08

Conclusion

The speed at which the second vaulter clears the bar is 1.67 m/s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Energy
Potential energy is the stored energy of position, held by an object due to its height above the ground. Imagine lifting a book; the book gains potential energy due to the gravitational field acting on it as it goes higher. The amount of potential energy can be calculated using the formula:
  • \[ PE = mgh \]
  • where \(m\) is the mass of the object, \(g\) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\) on Earth), and \(h\) is the height above the ground.
Potential energy plays a central role in problems involving conservation of energy. In the case of the pole-vaulter, as they reach the top of the jump, their potential energy is at its maximum. Understanding this concept helps us see how energy transfers in physical systems, like converting potential energy to kinetic energy as the vaulter falls back down.
Kinetic Energy
Kinetic energy is the energy of motion. Whenever an object moves, it possesses kinetic energy. The kinetic energy of an object is determined by its speed and mass using the formula:
  • \[ KE = \frac{1}{2} mv^2 \]
  • where \(m\) is mass and \(v\) is velocity.
During the exercise, kinetic energy is observed in the pole-vaulter both when clearing the bar and during landing. As potential energy decreases from the height due to gravity doing work, kinetic energy increases. This conversion demonstrates the principle of conservation of mechanical energy鈥攅nergy is not lost but transformed between forms. The vaulter's speed as they clear the bar and ultimately land is a function of how much kinetic energy they have gained from the conversion of potential energy.
Physics Problem Solving
Physics problem solving often involves applying fundamental principles, like the conservation of energy, to understand and predict the behavior of systems. The step-by-step method outlined in the exercise is a systematic approach:
  • Start by understanding the problem and identifying given values and required unknowns.
  • Use relevant physics formulas, like those for kinetic and potential energy, to express the conservation of energy.
  • Solve for unknowns using algebra, transforming and rearranging equations as needed.
For the pole-vaulter problem, solving required finding the initial speed using known landing speeds and heights derived from conservation of energy principles. This analytical approach helps break down complex problems into manageable parts, enabling clearer understanding and accurate solutions. Practicing structured problem solving ensures better grasp of physical concepts and fosters confidence in tackling similar challenges.

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Most popular questions from this chapter

The drawing shows two boxes resting on frictionless ramps. One box is relatively light and sits on a steep ramp. The other box is heavier and rests on a ramp that is less steep. The boxes are released from rest at \(\mathrm{A}\) and allowed to slide down the ramps. Which box, if either, has (a) the greater speed and (b) the greater kinetic energy at B? Provide a reason for each answer. The two boxes have masses of 11 and \(44 \mathrm{~kg}\). If \(\mathrm{A}\) and \(\mathrm{B}\) are 4.5 and \(1.5 \mathrm{~m},\) respectively, above the ground, determine the speed of (a) the lighter box and (b) the heavier box when each reaches \(\mathrm{B}\). (c) What is the ratio of the kinetic energy of the heavier box to that of the lighter box at B? Be sure that your answers are consistent with your answers to the Concept Questions.

A cable lifts a \(1200-\mathrm{kg}\) elevator at a constant velocity for a distance of \(35 \mathrm{~m}\). What is the work done by (a) the tension in the cable and (b) the elevator's weight?

A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force does work in raising the helicopter. (a) What type (s) of energy is (are) changing? Is each type increasing or decreasing? Why? (b) How is (are) the type(s) of energy that is (are) changing related to the work done by the lifting force? (c) If you want to determine the average power generated by the lifting force, what other variable besides the work must be known? An \(810-\mathrm{kg}\) helicopter rises from rest to a speed of \(7.0 \mathrm{~m} / \mathrm{s}\) in a time of \(3.5 \mathrm{~s}\). During this time it climbs to a height of \(8.2 \mathrm{~m}\). What is the average power generated by the lifting force?

A husband and wife take turns pulling their child in a wagon along a horizontal sidewalk. Each exerts a constant force and pulls the wagon through the same displacement. They do the same amount of work, but the husband's pulling force is directed \(58^{\circ}\) above the horizontal, and the wife's pulling force is directed \(38^{\circ}\) above the horizontal. The husband pulls with a force whose magnitude is \(67 \mathrm{~N}\). What is the magnitude of the pulling force exerted by his wife?

Relative to the ground, what is the gravitational potential energy of a 55.0 -kg person who is at the top of the Sears Tower, a height of \(443 \mathrm{~m}\) above the ground?
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