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Chapter 6: Problem 69

A pole-vaulter approaches the takeoff point at a speed of \(9.00 \mathrm{~m} / \mathrm{s}\). Assuming that only this speed determines the height to which he can rise, find the maximum height at which the vaulter can clear the bar.

Short Answer

Expert verified
The pole-vaulter can clear a maximum height of approximately 4.13 meters.

Step by step solution

01

Understand Energy Conversion

In this problem, the kinetic energy of the vaulter is converted into gravitational potential energy as the vaulter rises to his maximum height. The initial kinetic energy is given by \( KE = \frac{1}{2}mv^2 \) where \( m \) is the mass and \( v = 9.00 \mathrm{~m/s} \) (given). The gravitational potential energy at the maximum height is \( PE = mgh \), where \( h \) is the height and \( g = 9.81 \mathrm{~m/s}^2 \) is the acceleration due to gravity.
02

Set KE Equal to PE

At the maximum height, all the kinetic energy is converted into potential energy, so \( \frac{1}{2}mv^2 = mgh \). We notice that the mass \( m \) cancels out from both sides because it is present in both terms.
03

Solve for Maximum Height

After canceling \( m \), the equation simplifies to \( \frac{1}{2}v^2 = gh \). Rewriting this for \( h \), we get \( h = \frac{v^2}{2g} \).
04

Substitute Values and Calculate

Substitute the given speed \( v = 9.00 \mathrm{~m/s} \) and \( g = 9.81 \mathrm{~m/s}^2 \) into \( h = \frac{v^2}{2g} \). Therefore, \( h = \frac{(9.00)^2}{2 \times 9.81} \approx \frac{81}{19.62} \approx 4.13 \mathrm{~m} \). The maximum height the vaulter can clear is approximately \( 4.13 \) meters.

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses because of its position in a gravitational field. Imagine lifting an object in the air; you're working against gravity. This energy depends on three main factors:
  • The mass of the object (\( m \))
  • The height (\( h \)) above a reference point, usually the ground
  • The acceleration due to gravity (\( g \), approximately \( 9.81 \, \mathrm{m/s}^2 \) on Earth)
The formula for gravitational potential energy is \( PE = mgh \). As an object rises, its gravitational potential energy increases because it's gaining height. In the case of the pole-vaulter, as he vaults up, his speed gets converted into this potential energy, which determines how high up he can go before coming back down.
Energy Conversion
Energy conversion is a fundamental concept in physics. It refers to the process of changing one form of energy into another. In the example of a pole-vaulter, we see the conversion of kinetic energy to gravitational potential energy.Initially, the vaulter heads towards the takeoff point with a kinetic energy, calculated using the formula \( KE = \frac{1}{2}mv^2 \). As he rises, this kinetic energy gets converted into gravitational potential energy. At his maximum height, all the kinetic energy is turned into potential energy.This conversion is key because it helps in understanding how energy transformations determine motion and positions:
  • A moving car (kinetic energy) stops and the brakes get hot (thermal energy).
  • A cyclist speeds down a hill (kinetic energy) and then heads upwards, slowing down as energy changes into potential energy.
Through these types of conversion, we see how energy is never lost but transformed, enabling us to calculate outcomes like the maximum height cleared by a pole-vaulter.
Mechanics in Physics
Mechanics is a branch of physics dealing with the motion and forces acting on objects. It's divided into two main areas: kinematics, which describes motion without regard to causes, and dynamics, which explains motion through forces. In the context of the pole-vaulter:
  • Kinematics: Describes how fast the vaulter runs and how gradually he rises and falls.
  • Dynamics: Explains how forces such as gravity influence his motion and help us understand the energy transformations.
Mechanics principles such as Newton's laws of motion play a critical role. For instance, Newton's first law states an object in motion stays in motion unless acted on by an external force. This is seen when the vaulter continues moving upwards until gravity (a force) causes him to decelerate, reach his peak, and descend. Through mechanics, we thoroughly examine motion. By understanding these principles, it becomes easier to solve problems like this exercise, where forces and energy conversions are analyzed to determine outcomes like the maximum achievable height.

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Most popular questions from this chapter

A fighter jet is launched from an aircraft carrier with the aid of its own engines and a steam-powered catapult. The thrust of its engines is \(2.3 \times 10^{5} \mathrm{~N}\). In being launched from rest it moves through a distance of \(87 \mathrm{~m}\) and has a kinetic energy of \(4.5 \times 10^{7} \mathrm{~J}\) at liftoff. What is the work done on the jet by the catapult?

A pitcher throws a \(0.140\) -kg baseball, and it approaches the bat at a speed of \(40.0 \mathrm{~m} / \mathrm{s}\). The bat does \(W_{n c}=70.0 \mathrm{~J}\) of work on the ball in hitting it. Ignoring air resistance, determine the speed of the ball after the ball leaves the bat and is \(25.0 \mathrm{~m}\) above the point of impact.

When a \(0.045-\mathrm{kg}\) golf ball takes off after being hit, its speed is \(41 \mathrm{~m} / \mathrm{s}\). (a) How much work is done on the ball by the club? (b) Assume that the force of the golf club acts parallel to the motion of the ball and that the club is in contact with the ball for a distance of \(0.010 \mathrm{~m}\). Ignore the weight of the ball and determine the average force applied to the ball by the club.

A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. (a) Does the kinetic frictional force do positive, negative, or zero work? Provide a reason for your answer. (b) Does the total mechanical energy of the student increase, decrease, or remain the same as she descends the slide? Why? (c) If the kinetic frictional force does work, how is this work related to the change in the total mechanical energy of the student? The student has a mass of \(83.0 \mathrm{~kg}\) and the height of the water slide is \(11.8 \mathrm{~m}\). If the kinetic frictional force does \(-6.50 \times 10^{3} \mathrm{~J}\) of work, how fast is the student going at the bottom of the slide?

Refer to Interactive Solution \(6.51\) for a review of the approach taken in problems such as this one. A \(67.0\) -kg person jumps from rest off a \(3.00\) -m-high tower straight down into the water. Neglect air resistance during the descent. She comes to rest \(1.10 \mathrm{~m}\) under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is nonconservative.
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