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If a 70 kg person climbs 10 flights of stairs, each flight \(3 \mathrm{~m}\) high, how much potential energy have they gained?

Short Answer

Expert verified
Answer: The person has gained 20,574 Joules of potential energy after climbing 10 flights of stairs, each with a height of 3 meters.

Step by step solution

01

Identify the given information

From the problem, we are given the following information: - Mass of the person (\(m\)): \(70 \mathrm{~kg}\) - Height of each flight of stairs (\(h_1\)): \(3 \mathrm{~m}\) - Number of flights of stairs climbed (\(n\)): \(10\)
02

Calculate the total height climbed

To calculate the total height climbed, we need to multiply the height of each flight of stairs by the number of flights climbed: Total height (\(h\)) = height of each flight (\(h_1\)) × number of flights (\(n\)) In our case: \(h\) = \(3 \mathrm{~m}\) × \(10\) = \(30 \mathrm{~m}\).
03

Calculate the potential energy gained

Now, we can calculate the potential energy using the formula: \(PE = mgh\). We will use the following values: - Mass of the person (\(m\)): \(70 \mathrm{~kg}\) - Gravitational acceleration (\(g\)): \(9.81 \mathrm{m/s^2}\) - Total height climbed (\(h\)): \(30 \mathrm{~m}\) Plugging in the values, we get: \(PE = (70 \mathrm{~kg}) \times (9.81 \mathrm{m/s^2}) \times (30 \mathrm{~m})\)
04

Solve for potential energy

Finally, we will solve for the potential energy: \(PE = 70 \times 9.81 \times 30\) \(PE = 20574 \mathrm{~J}\) (Joules) The person has gained \(20,574\) Joules of potential energy after climbing \(10\) flights of stairs, each with a height of \(3 \mathrm{~m}\).

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Key Concepts

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

Gravitational Potential Energy
When you lift an object against the force of gravity, you store energy in that object. This stored energy is known as Gravitational Potential Energy (GPE). It is a type of potential energy that depends on the height of an object above the ground. The higher you lift the object, the more potential energy it gains. This energy can be calculated using the formula: \[ PE = mgh \] Where:
  • \(m\) is the mass of the object in kilograms (kg)
  • \(g\) is the acceleration due to gravity, approximately \(9.81 \text{ m/s}^2\) on Earth
  • \(h\) is the height in meters (m) above the reference point
In our example, a person climbing stairs gains potential energy as they rise higher. The energy is proportional to the mass of the person, the height climbed, and the gravitational acceleration.
Physics Calculations
Physics calculations involve a series of logical steps to find an answer to a problem. Understanding the process behind these calculations is essential: To calculate Gravitational Potential Energy in the exercise, the first step is to gather all the relevant data from the problem, such as mass, gravitational acceleration, and height. Next, it's crucial to use the correct formulas and consistently apply units throughout the calculation.

Step-by-Step Calculation Process

  • **Identify given information**: Mass \(m = 70\, kg\) and total height \(h = 30\, m\)
  • **Use the formula**: \[ PE = mgh \]
  • **Apply values**: Insert \(m = 70\, kg\), \(g = 9.81\, m/s^2\), and \(h = 30\, m\) into \[ PE = 70 \times 9.81 \times 30 \]
  • **Calculate**: Perform the multiplication to find \(PE = 20574\, J\) (Joules)
The step-by-step approach helps ensure accuracy and can be applied to various physics problems.
Energy Conservation
Energy conservation is a fundamental principle in physics, stating that energy cannot be created or destroyed, only transformed from one form to another. When a person climbs stairs, chemical energy from food is converted into mechanical energy, which is then converted into Gravitational Potential Energy as they elevate.
This principle is key for understanding how energy flows and converts during different activities. For instance, when the person descends, the potential energy can transform into kinetic energy as they speed downward, continuing to obey the law of conservation.

Applications of Energy Conservation

  • Used in understanding mechanisms of various physical phenomena
  • Helps in solving complex problems by accounting for all energy types
  • Integral for the design of machines and systems that efficiently manage energy use
Thus, acknowledging energy conservation provides a comprehensive view of energy transformations in all processes including everyday actions like climbing stairs.

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Most popular questions from this chapter

A \(10 \mathrm{~kg}\) box is lifted \(2 \mathrm{~m}\) off the floor and placed on a frictionless horizontal conveyor to take it \(30 \mathrm{~m}\) across a warehouse. At the end of the conveyor, it is lowered \(1 \mathrm{~m}\) where it ends up on a shelf. How much net gravitational potential energy was given to the box from the start to the end of the process?

Fig. \(10.1\) (p. 167 ) indicated that about 44,000 TW globally goes into evaporating water. We can turn this into an estimate of how much rain we expect per year, on average. The simplest way to do this is to think of a single square meter of ocean surface, receiving an average evaporation input power of \(120 \mathrm{~W} .^{44}\) Each millimeter of of water depth across our square meter has a volume of \(1 \mathrm{~L}\), or a mass of \(1 \mathrm{~kg}\). At a steady input of \(120 \mathrm{~W}\), \(^{45}\) how many millimeters of water are drawn off in a year? That same amount will come back down somewhere as precipitation.

A typical American household uses approximately \(30 \mathrm{kWh}\) per day of electricity. Convert this to Joules and then imagine building a water tank \(10.8 \mathrm{~m}\) above the house \(^{37}\) to supply one day's worth of electricity. \(^{38}\) How much mass of water is this, in kg? At a density of \(1,000 \mathrm{~kg} / \mathrm{m}^{3}\), what is the volume in cubic meters, and what is the side length of a cube \(^{39}\) having this volume? Take a moment to visualize (or sketch) this arrangement.

A dam 50 meters high is constructed on a river and is delivering \(180 \mathrm{MW}\) at some moment in time. What is the flow rate of water, in cubic meters per second, if the facility converts gravitational potential energy into electricity at \(90 \%\) efficiency?

While the Chief Joseph Dam on the Columbia River can generate as much as \(2.62 \mathrm{GW}\left(2.62 \times 10^{9} \mathrm{~W}\right)\) of power at full flow, the river does not always run at full flow. The average annual production is 10.7 TWh per year \(\left(10.7 \times 10^{12} \mathrm{Wh} / \mathrm{yr}\right)\). What is the capacity factor of the dam: the amount delivered vs. the amount if running at \(100 \%\) capacity the whole year?

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