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

Short Answer

Expert verified
The second vaulter clears the bar at a speed of approximately 1.34 m/s.

Step by step solution

01

Understand the Concept

The problem involves the application of conservation of energy principles. We will equate the sum of potential and kinetic energy at the top of the jump with the kinetic energy as they land on the ground.
02

Formula for Energy Conservation

Using the energy conservation principle, the total mechanical energy at the top is equal to the total mechanical energy when landing: Initial Energy (top) = Kinetic Energy (landing)For the first vaulter: \[ \frac{1}{2} m v_{1}^{2} + mgh = \frac{1}{2} m (8.90)^2 \]For the second vaulter: \[ \frac{1}{2} m v_{2}^{2} + mgh = \frac{1}{2} m (9.00)^2 \]
03

Simplifying Equations

Since both vaulters clear the bar at the same height, the potential energy expression \(mgh\) is the same for both: - First vaulter clearance: \[ \frac{1}{2} (1.00)^2 + gh = \frac{1}{2} (8.90)^2 \]- Second vaulter: \[ \frac{1}{2} v_{2}^{2} + gh = \frac{1}{2} (9.00)^2 \] Since \(gh\) is the same, we can find \(gh\) from the first equation and use it in the second.
04

Solving for \(gh\)

Re-arrange the first vaulter's equation to solve for \(gh\): \[ gh = \frac{1}{2} \times 8.90^2 - \frac{1}{2} \times 1.00^2 \]Calculate:\[ gh = \frac{1}{2} (79.21) - 0.5 = 39.605 \text{ m}^2/\text{s}^2 \]
05

Calculate \(v_2\) for the second vaulter

Substitute \(gh\) into the second vaulter's equation to solve for \(v_2\): \[ \frac{1}{2} v_{2}^{2} + 39.605 = \frac{1}{2} (9.00)^2 \]Simplify:\[ \frac{1}{2} v_{2}^{2} = \frac{1}{2} (81) - 39.605 \]Solve for \(v_{2}^{2}\): \[ v_{2}^{2} = 81 - 2 \times 39.605 \]\[ v_{2}^{2} = 1.79 \]Find \(v_2\):\[ v_2 = \sqrt{1.79} \approx 1.34 \text{ m/s} \]
06

Verify Your Solution

Check the calculations and ensure all steps followed logical progression. Verify consistency with principles stated, e.g., speeds and potential energy equality. Result seems consistent with principles of energy conservation.

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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 the energy an object possesses due to its motion. You can think of it as the energy that helps move things. Every moving object has kinetic energy, which depends on how fast the object moves and its mass.

For any object, including our pole-vaulters, the formula to calculate kinetic energy is: \[ KE = \frac{1}{2} mv^2 \]where:
  • \( m \) is the mass of the object
  • \( v \) is the velocity of the object
The faster something moves, the more kinetic energy it has. This is why both pole-vaulters have different kinetic energies when they land; despite landing at a slightly different speeds, the minor change in speed makes a difference in their kinetic energy.

It’s important to note in problems like these, mass \( m \) often cancels out because both energy states consider the same object. Thus, the changes in speed beautifully showcase the shift in energy forms.
Potential Energy
Potential energy is the stored energy an object has due to its position or height. When you raise an object higher up, you add potential energy to it because of gravity's pull.

Consider the formula for gravitational potential energy:\[ PE = mgh \]where:
  • \( m \) is the mass of the object
  • \( g \) is the acceleration due to gravity (approximately \( 9.81 \text{ m/s}^2 \) on Earth)
  • \( h \) is the height above the ground
In the case of our pole-vaulters, both have stored a similar amount of potential energy because they cleared the bar at the same height. This potential energy becomes crucial in understanding how it converts to kinetic energy as they descend.

When the vaulters clear the bar and fall back to the ground, this potential energy transforms into kinetic energy. The key point here shows how potential energy at the top of the jump converts to increased speeds as they reach the ground.
Mechanical Energy
Mechanical energy combines an object's kinetic and potential energy. It helps in understanding how energy transforms between different forms during motion.

The principle of conservation of mechanical energy states that in the absence of non-conservative forces like friction, the total mechanical energy of a system remains constant. This means:\[ ME_{initial} = ME_{final} \]or\[ KE_{initial} + PE_{initial} = KE_{final} + PE_{final} \]

For our pole-vaulters, this principle ensures their total energy remains the same from start to finish of their jumps. At the top, where they barely clear the bar, a significant portion of their energy is stored as potential energy. When they land, this energy shifts almost entirely to kinetic, as potential energy decreases when they lose height.

This conversion is what makes the conservation of energy such a powerful tool. By knowing how fast each vaulter lands, we can backtrack to find their speed over the bar, showing how initial positions dictate the final outcomes, which were solved in our step-by-step analysis of the second vaulter's situation.

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Most popular questions from this chapter

A 55.0-kg skateboarder starts out with a speed of 1.80 m/s. He does 80.0 J of work on himself by pushing with his feet against the ground. In addition, friction does 265 J of work on him. In both cases, the forces doing the work are non conservative. The final speed of the skateboarder is 6.00 \(\mathrm{m} / \mathrm{s}\) . (a) Calculate the change \(\left(\Delta PE=PE_{\mathrm{f}}-PE_{0}\right)\) in the gravitational potential energy. (b) How much has the vertical height of the skater changed, and is the skater above or below the starting point?

A \(1200-\mathrm{kg}\) car is being driven up a \(5.0^{\circ}\) hill. The frictional force is directed opposite to the motion of the car and has a magnitude of \(f=524 \mathrm{N}\) . A force \(\overrightarrow{\mathbf{F}}\) is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight \(\overrightarrow{\mathbf{W}}\) and the normal force \(\overrightarrow{\mathbf{F}}_{\mathrm{N}}\) directed perpendicular to the road surface. The length of the road up the hill is 290 \(\mathrm{m}\) . What should be the magnitude of \(\overrightarrow{\mathbf{F}},\) so that the net work done by all the forces acting on the car is \(+150 \mathrm{kJ}\) ?

A 16-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 24 N. Starting from rest, the sled attains a speed of 2.0 m/s in 8.0 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.

The (non conservative) force propelling a \(1.50 \times 10^{3}-kg\) car up a mountain road does \(4.70 \times 10^{6} J\) of work on the car. The car starts from rest at sea level and has a speed of 27.0 \(m /s\) sat an altitude of \(2.00 \times 10^{2} m\) above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are non conservative forces.

At a carnival, you can try to ring a bell by striking a target with a 9.00-kg hammer. In response, a 0.400-kg metal piece is sent upward toward the bell, which is 5.00 m above. Suppose that 25.0% of the hammer’s kinetic energy is used to do the work of sending the metal piece upward. How fast must the hammer be moving when it strikes the target so that the bell just barely rings?

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