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How much more gravitational potential energy does a 1.0-kg hammer have when it is on a shelf \(1.2 \mathrm{~m}\) high than when it is on a shelf \(0.90 \mathrm{~m}\) high?

Short Answer

Expert verified
The hammer has 2.94 Joules more gravitational potential energy at 1.2 m height compared to 0.9 m.

Step by step solution

01

Understand the Gravitational Potential Energy Formula

Gravitational potential energy (GPE) is calculated using the formula: \(E_p = mgh\) where \(E_p\) is the gravitational potential energy, \(m\) is the mass of the object, \(g\) is the acceleration due to gravity (approximately 9.8 m/s²), and \(h\) is the height above the ground. We need to determine the difference in the GPE at two different heights.
02

Calculate GPE at 1.2 m Height

For the shelf at 1.2 m, substitute the values into the formula: \(m = 1.0 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), \(h = 1.2 \text{ m}\). Calculate: \(E_{p1} = 1.0 \times 9.8 \times 1.2\). This gives \(E_{p1} = 11.76\) Joules.
03

Calculate GPE at 0.9 m Height

For the shelf at 0.9 m, substitute the values into the formula: \(m = 1.0 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), \(h = 0.9 \text{ m}\). Calculate: \(E_{p2} = 1.0 \times 9.8 \times 0.9\). This gives \(E_{p2} = 8.82\) Joules.
04

Find the Difference in GPE

Subtract the gravitational potential energy at 0.9 m from the gravitational potential energy at 1.2 m: \(\Delta E_p = E_{p1} - E_{p2} = 11.76 - 8.82\). This gives \(\Delta E_p = 2.94\) Joules.

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Key Concepts

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

Physics Education
Physics education is a wonderful journey into understanding how the universe operates by exploring fundamental concepts. One such concept is energy, specifically gravitational potential energy. Teaching about energy in physics often involves connecting principles to real-world situations, like the exercise involving the hammer. This exercise allows students to apply formulas to calculate changes in gravitational potential energy as objects change their height.

By using engaging examples, students can appreciate how physical laws apply in everyday scenarios. Solving these problems supports critical thinking and strengthens problem-solving skills. Incorporating physics problems into education helps students grasp abstract concepts by making them tangible. Exercises like these are not just equations; they offer insights into the world around us and how energy plays a pivotal role in different circumstances.
  • Connects theory with real-world applications
  • Strengthens problem-solving and critical thinking skills
  • Makes abstract concepts tangible
Energy Calculation
Energy calculation is an essential skill in physics, letting us understand the dynamics and interactions of objects. In our exercise, we compute the gravitational potential energy of a hammer based on its height. Here, we see how energy changes with position, which is crucial for predicting how objects move. Calculating energy involves using known values like mass, height, and gravitational acceleration.

These calculations often build on straightforward mathematical operations: multiplication and subtraction. We can simplify the process into clear steps:
  • Identify the variables: mass, gravitational acceleration, and height
  • Apply the gravitational potential energy formula to each scenario
  • Compute the energies at different heights using given values
  • Determine the energy difference to find results
Regular practice with these steps enhances students' understanding of energy transformations in physics and prepares them for more advanced applications such as mechanical systems and engineering problems.
Gravitational Potential Energy Formula
The gravitational potential energy formula is instrumental in calculating the energy stored due to an object’s height above ground level. Mathematically, the formula is expressed as: \[ E_p = mgh \]where \(E_p\) is the gravitational potential energy, \(m\) is the object's mass, \(g\) is the acceleration due to gravity (approximated as 9.8 m/s² on Earth), and \(h\) is the height relative to a reference point.

This formula shows how energy correlates directly with both height and mass. In our exercise, we're interested in the energy difference between two heights: 1.2 m and 0.9 m.

Through calculating \(E_{p1} = mgh\) and \(E_{p2} = mgh\) for each height and finding the difference \(\Delta E_p = E_{p1} - E_{p2}\), we reveal how much more energy an object holds at a greater height. This insight is not only critical for understanding potential energy but also for exploring concepts like work done by gravity, conservation of energy, and how energy is conserved as it transitions between different forms.
  • Highlights relationship between height, mass, and energy
  • Demonstrates conservation and transformation of energy
  • Serves as a foundation for more complex physics concepts

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Most popular questions from this chapter

The stopping distance of a vehicle is an important safety factor. Assuming a constant braking force, use the work-energy theorem to show that a vehicle's stopping distance is proportional to the square of its initial speed. If an automobile traveling at \(45 \mathrm{~km} / \mathrm{h}\) is brought to a stop in \(50 \mathrm{~m}\), what would be the stopping distance for an initial speed of \(90 \mathrm{~km} / \mathrm{h} ?\)

IE A 0.455 -kg soccer ball is kicked off level ground at an angle of \(40^{\circ}\) with an initial speed of \(30.0 \mathrm{~m} / \mathrm{s}\). Neglecting air resistance, (a) at its maximum height off the ground, its kinetic energy will be (1) less than its value at launch, but not zero, (2) more than its value at launch, (3) zero. Explain. (b) Determine its kinetic energy when it is at its maximum height above the ground and compare it to the kinetic energy at launch. [Hint: What is its velocity at the top of its arc? Review projectile motion if necessary.]

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A pump lifts \(200 \mathrm{~kg}\) of water per hour a height of \(5.0 \mathrm{~m} .\) What is the minimum necessary power output rating of the water pump in watts and horsepower?

If the work required to speed up a car from \(10 \mathrm{~km} / \mathrm{h}\) to \(20 \mathrm{~km} / \mathrm{h}\) is \(5.0 \times 10^{3} \mathrm{~J},\) what would be the work required to increase the car's speed from \(20 \mathrm{~km} / \mathrm{h}\) to \(30 \mathrm{~km} / \mathrm{h} ?\)

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