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A pole-vaulter just clears the bar at \(5.80 \mathrm{m}\) and falls back to the ground. The change in the vaulter's potential energy during the fall is \(-3.70 \times 10^{3} \mathrm{J} .\) What is his weight?

Short Answer

Expert verified
The vaulter's weight is approximately 638 N.

Step by step solution

01

Identify formula for potential energy change

The change in gravitational potential energy is given by \[ \Delta U = mgh_{f} - mgh_{i} \]. Since the vaulter falls back to the ground, the final height \(h_{f}\) is 0. Therefore, the change in potential energy is \( \Delta U = -mgh_{i} \), where \(m\) is the mass, \(g\) is the gravity (\(9.8 \, \text{m/s}^2\)), and \(h_{i}\) is the initial height.
02

Rearrange to find weight (\(mg\))

We can rearrange the formula for change in potential energy to find the weight. The formula becomes: \[ -mgh_{i} = \Delta U \]. Simplifying further gives us \[ mg = -\frac{\Delta U}{h_{i}} \]. In this formula, \(mg\) represents the weight.
03

Substitute known values

Now, substitute the given values into the rearranged equation: \[ mg = -\frac{-3.70 \times 10^{3} \, \text{J}}{5.80 \, \text{m}} \]. This is because the change in potential energy during the fall is \(-3.70 \times 10^{3} \, \text{J}\) and the initial height \(h_i\) is \(5.80 \, \text{m}\).
04

Calculate the weight

Perform the division to find the weight: \[ mg = \frac{3.70 \times 10^{3}}{5.80} \approx 637.93 \, \text{N} \]. Thus, the weight of the pole vaulter is approximately \(638 \, \text{N}\) when rounded to the nearest whole number.

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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 (GPE) is the energy possessed by an object due to its position in a gravitational field. When an object is elevated above the ground, it has the potential to do work as it falls. This form of energy is dependent on three factors: the object's mass (\(m\)), gravitational acceleration (\(g\)), which is usually \(9.8 \, \text{m/s}^2\), the standard on Earth, and the height (\(h\)) from which it falls.
The formula to calculate gravitational potential energy is \( U = mgh \). This equation shows that the energy is directly proportional to both the mass of the object and its height above the ground.
  • A greater mass or height results in more gravitational potential energy.
  • If an object falls to the ground, its GPE decreases as it loses height.
Understanding GPE is crucial in physics, especially in problems concerning free-falling objects. For a pole-vaulter clearing a bar, as in the example problem, their GPE changes as they reach the apex and descend to the ground.
Weight Calculation
Calculating weight is an essential step in solving physics problems involving potential energy. Weight is simply the force exerted by gravity on an object, and it is denoted by \( mg \). In physics, weight is a vector quantity, meaning it has both magnitude and direction.
The formula \( mg = -\frac{\Delta U}{h_i} \) is derived from the equation for gravitational potential energy change. Here, \( \Delta U \) is the change in potential energy, and \( h_i \) is the initial height. By rearranging this formula, we can solve for \( mg \), the weight of the object.
  • This approach is incredibly useful when given changes in potential energy and height.
  • Weight plays a crucial role in understanding how much force is necessary to keep an object at rest in a gravity field.
By applying the values from the original problem, we find that the pole vaulter's weight is approximately \( 638 \, \text{N} \), emphasizing the consistent role of gravity.
Pole Vaulting
Pole vaulting is a fascinating sport and an excellent example of physics principles at work. It involves an athlete running with a pole, planting it into the ground, and using it to vault over a high bar. This action is more than just skill; it’s a perfect demonstration of potential and kinetic energy exchange.
At the start, the athlete's kinetic energy, gained from running, converts into gravitational potential energy as they ascend. Once they clear the bar, this potential energy converts back into kinetic energy as they fall to the ground.
  • The higher the vaulter goes, the more GPE they possess at the peak of their jump.
  • Understanding these energy transformations is crucial for enhancing performance and preventing injuries.
This intricate balance between energies showcases the essence of energy conservation—a fundamental concept in physics.
Physics Problem Solving
Physics problem solving is a critical skill that requires understanding and applying various formulas and concepts to find a solution. When faced with a problem, like determining the weight of a pole-vaulter, it's essential to follow a logical step-by-step process:

First, identify the core physics concepts involved, such as potential energy in this case. Then, apply the relevant formulas, carefully considering each variable and what it's representing in the scenario. Rearrange the equations to find the unknown, substituting given values carefully.
  • Always cross-check your answers to ensure they make sense in real-world contexts.
  • This systematic approach helps in tackling a wide range of physics problems effectively.
By practicing these steps, one can build confidence and proficiency in solving complex physics questions, turning theoretical understanding into practical skill.

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Most popular questions from this chapter

A car accelerates uniformly from rest to \(20.0 \mathrm{m} / \mathrm{s}\) in \(5.6 \mathrm{s}\) along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is \(9.0 \times 10^{3} \mathrm{N}\) and (b) the weight of the car is \(1.4 \times 10^{4} \mathrm{N}\).

A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. A \(70.3-\mathrm{kg}\) water-skier has an initial speed of \(6.10 \mathrm{m} / \mathrm{s}\). Later, the speed increases to \(11.3 \mathrm{m} / \mathrm{s} .\) Determine the work done by the net external force acting on the skier.

A person pulls a toboggan for a distance of \(35.0 \mathrm{m}\) along the snow with a rope directed \(25.0^{\circ}\) above the snow. The tension in the rope is 94.0 N. (a) How much work is done on the toboggan by the tension force? (b) How much work is done if the same tension is directed parallel to the snow?

Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilo-gram generated by seven-time-winner Lance Armstrong \((m=75.0 \mathrm{kg})\) is \(6.50 \mathrm{W}\) per kilogram of his body mass. (a) How much work does he do during a \(135-\mathrm{km}\) race in which his average speed is \(12.0 \mathrm{m} / \mathrm{s} ?\) (b) Often, the work done is expressed in nutritional Calories rather than in joules. Express the work done in part (a) in terms of nutritional Calories, noting that 1 joule \(=2.389 \times 10^{-4}\) nutritional Calories.

6\. A person pushes a 16.0-kg shopping cart at a constant velocity for a distance of \(22.0 \mathrm{m} .\) She pushes in a direction \(29.0^{\circ}\) below the horizontal. A \(48.0-N\) frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts? Determine the work done by (b) the pushing force, (c) the frictional force, and (d) the gravitational force.

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