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Ignoring details associated with friction, extra forces exerted by arm and leg muscles, and other factors, we can consider a pole vault as the conversion of an athlete's running kinetic energy to gravitational potential energy. If an athlete is to lift his body \(4.8 \mathrm{m}\) during a vault, what speed must he have when he plants his pole?

Short Answer

Expert verified
The athlete must have a speed of about \(9.68 m/s\) when he plants his pole to lift his body \(4.8 m\) during the vault.

Step by step solution

01

Identify the given information and the unknown value

The given information is that the athlete needs to lift his body height by \(4.8 m\). We need to find the speed (\(v\)) he must have when he plants his pole.
02

Analyzing energy conversion

During the vault, the athlete's kinetic energy gets converted into gravitational potential energy. We can represent this energy conversion using the following formula: \[K_{initial} = U_{final}\] Here, \(K_{initial}\) is the initial kinetic energy, while \(U_{final}\) is the final gravitational potential energy.
03

Calculate the initial kinetic energy

The initial kinetic energy can be calculated using the formula: \[K_{initial} = \frac{1}{2}mv^2\] Here, \(m\) is the mass of the athlete, and \(v\) is the speed the athlete must have when he plants his pole. We are looking for the value of \(v\).
04

Calculate the final gravitational potential energy

The final gravitational potential energy can be calculated using the formula: \[U_{final} = mgh\] Here, \(m\) is the mass of the athlete, \(g\) is the acceleration due to gravity (\(9.8 m/s^2\)), and \(h\) is the height the athlete lifts his body (\(4.8 m\)).
05

Equate the initial kinetic energy and the final gravitational potential energy

Now that we have the formulas for the initial kinetic energy and the final gravitational potential energy, we can equate them: \[\frac{1}{2}mv^2 = mgh\]
06

Solve for the speed, \(v\)

Notice that the mass of the athlete, \(m\), appears on both sides of the equation. We can divide both sides of the equation by \(m\) to obtain: \[\frac{1}{2}v^2 = gh\] Now, let's plug in the given values for \(g\) and \(h\): \[\frac{1}{2}v^2 = 9.8 \cdot 4.8\] Next, multiply both sides by 2: \[v^2 = 2 \cdot 9.8 \cdot 4.8\] Now, take the square root of both sides to solve for \(v\): \[v = \sqrt{2 \cdot 9.8 \cdot 4.8}\] Finally, calculate the value of \(v\): \[v \approx 9.68 m/s\] The athlete must have a speed of about \(9.68 m/s\) when he plants his pole to lift his body \(4.8 m\) during the vault.

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Key Concepts

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

Kinetic Energy
Kinetic energy is a key concept in the world of physics, particularly when tackling problems involving movement and speed. It represents the energy an object possesses due to its motion. Simply put, the faster something moves, the more kinetic energy it carries. This is crucial for athletes like pole vaulters, who rely on their running speed to generate this energy.

To find the kinetic energy, we use the formula:
  • \[K = \frac{1}{2}mv^2\]
Here, \(K\) is the kinetic energy, \(m\) is the mass of the object, and \(v\) is the velocity or speed. Notice that velocity is squared, indicating its significant impact on the energy value.

In the context of a pole vaulter, before they plant their pole , their running speed translates into kinetic energy. This energy is then converted into other forms, such as gravitational potential energy, as they lift off the ground.

Understanding kinetic energy helps in optimizing performance. Athletes must achieve a balance between their speed and the technique used, ensuring that their kinetic energy efficiently converts to the desired height.
Gravitational Potential Energy
Gravitational potential energy is the energy stored in an object due to its position relative to a gravitational field, like that of Earth. This form of energy increases with height; thus, the higher an object is elevated, the greater its potential energy becomes.

The formula to calculate gravitational potential energy is:
  • \[U = mgh\]
In this equation, \(U\) is the gravitational potential energy, \(m\) is the mass of the object, \(g\) is the acceleration due to gravity (approximately \(9.8 \, m/s^2\) on Earth), and \(h\) is the height above the ground.

In our pole vault example, the athlete's goal is to maximize their height. Therefore, their initial kinetic energy needs to be completely transformed into gravitational potential energy. When they reach the highest point of their vault, all of their speed from running is converted, lifting them to their desired height. This transformation underlines the essential energy shift from kinetic to potential during the athlete's vault.
Physics Problem Solving
In physics problem solving, it's all about understanding the relationship between physical quantities and effectively applying mathematical principles. Problems like the pole vault example often involve energy conversion, where solving them requires clear steps and systematic thinking.

Here's a typical way to approach such problems:
  • **Understand the Concept:** Identify which physics principles are at play, like energy conservation in this case.
  • **Identify Known Values:** Gather all given data and express them using formulas.
  • **Utilize Equations:** Apply appropriate formulas to connect known values with unknown ones.
  • **Solve the Problem:** Carefully work through the mathematics, ensuring to check unit consistency.
In our solution process, we equated kinetic and potential energy to find the required speed. Simplifying the process by canceling the athlete's mass on both sides of the equation. Physical problems often require understanding that some terms may repeat and can be simplified, much like the division of \(m\) on both sides in the equation \(\frac{1}{2}mv^2 = mgh\).

The art of physics problem solving lies not just in understanding formulas but seeing through complex relationships, generating insights, and accurately arriving at conclusions.

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Most popular questions from this chapter

A 1.0 -kg ball at the end of a 2.0 -m string swings in a vertical plane. At its lowest point the ball is moving with a speed of \(10 \mathrm{m} / \mathrm{s}\). (a) What is its speed at the top of its path? (b) What is the tension in the string when the ball is at the bottom and at the top of its path?

A force \(F(x)=\left(-5.0 x^{2}+7.0 x\right) \mathrm{N}\) acts on a particle. (a) How much work does the force do on the particle as it moves from \(x=2.0 \mathrm{m}\) to \(x=5.0 \mathrm{m} ?\) (b) Picking a convenient reference point of the potential energy to be zero at \(x=\infty,\) find the potential energy for this force.

An object of mass \(10 \mathrm{kg}\) is released at point \(A,\) slides to the bottom of the \(30^{\circ}\) incline, then collides with a horizontal massless spring, compressing it a maximum distance of \(0.75 \mathrm{m}\). (See below.) The spring constant is 500 \(\mathrm{M} / \mathrm{m},\) the height of the incline is \(2.0 \mathrm{m},\) and the horizontal surface is frictionless. (a) What is the speed of the object at the bottom of the incline? (b) What is the work of friction on the object while it is on the incline? (c) The spring recoils and sends the object back toward the incline. What is the speed of the object when it reaches the base of the incline? (d) What vertical distance does it move back up the incline?

Tarzan grabs a vine hanging vertically from a tall tree when he is running at \(9.0 \mathrm{m} / \mathrm{s}\). (a) How high can he swing upward? (b) Does the length of the vine affect this height?

A 300 g hockey puck is shot across an ice-covered pond. Before the hockey puck was hit, the puck was at rest. After the hit, the puck has a speed of 40 m/s. The puck comes to rest after going a distance of \(30 \mathrm{m}\). (a) Describe how the energy of the puck changes over time, giving the numerical values of any work or energy involved. (b) Find the magnitude of the net friction force.

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