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Chapter 7: Problem 1

In one day, a 75kg mountain climber ascends from the 1500m level on a vertical cliff to the top at 2400 m. The next day, she descends from the top to the base of the cliff, which is at an elevation of 1350 m. What is her change in gravitational potential energy (a) on the first day and (b) on the second day?

Short Answer

Expert verified
First day: 662175 J; Second day: -772875 J.

Step by step solution

01

Understand Gravitational Potential Energy

Gravitational potential energy (GPE) is given by the formula \( PE = mgh \), where \( m \) is mass, \( g \) is acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) on Earth), and \( h \) is height above a reference point. We need to calculate it for each specific change in altitude.
02

Calculate GPE on the First Day

On the first day, the change in GPE is determined by the difference in the initial and final heights. Use \( h_1 = 1500 \, \text{m} \) and \( h_2 = 2400 \, \text{m} \) for calculations. Thus, \( \Delta h = 2400 - 1500 = 900 \, \text{m} \). The GPE change is \( \Delta PE = mg \Delta h = 75 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 900 \, \text{m} \).
03

Compute the Energy Increase for the First Day

Substitute the values into the equation: \( \Delta PE = 75 \times 9.81 \times 900 = 662175 \, \text{J} \). Therefore, the change in gravitational potential energy on the first day is 662175 Joules.
04

Calculate GPE on the Second Day

On the second day, the climber descends. The initial height \( h_1 \) is 2400 m, and the final height \( h_2 \) is 1350 m. Calculate the change in height: \( \Delta h = 1350 - 2400 = -1050 \, \text{m} \). The GPE change is \( \Delta PE = mg \Delta h = 75 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times (-1050) \, \text{m} \).
05

Compute the Energy Decrease for the Second Day

Substitute the values into the equation: \( \Delta PE = 75 \times 9.81 \times (-1050) = -772875 \, \text{J} \). Therefore, the change in gravitational potential energy on the second day is -772875 Joules.

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Key Concepts

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

Energy Calculation
Energy calculation is an important aspect in physics, especially when dealing with gravitational potential energy. To calculate gravitational potential energy (GPE), the formula used is \( PE = mgh \), where:
  • \( m \) is the mass of the object in kilograms (kg).
  • \( g \) represents the acceleration due to gravity, which is approximately \( 9.81 \, \text{m/s}^2 \) on Earth.
  • \( h \) is the height in meters (m) above a reference point.
To find the change in GPE, we can look at the change in height \( \Delta h \). The formula becomes \( \Delta PE = mg \Delta h \). When calculating energy changes, remember:
  • Positive \( \Delta h \) indicates an increase in height, resulting in an increase in GPE.
  • Negative \( \Delta h \) indicates a decrease in height, corresponding to a decrease in GPE.
For example, in the case of our mountain climber, calculating energy changes on different days involves substituting these values into the formula for each segment of the journey.
Physics Problem Solving
Physics problem solving often involves breaking down a problem into manageable steps. By doing so, we can systematically approach and solve complex issues, such as calculating changes in gravitational potential energy. Here鈥檚 a simple strategy to navigate through such problems:
  • **Understand the Concept**: Clearly identify what you are asked to find. In this case, it鈥檚 the change in gravitational potential energy.
  • **Identify Known Variables**: Establish the necessary values, such as mass, height changes, and the acceleration due to gravity.
  • **Apply the Right Formula**: Use the appropriate formula, like \( PE = mgh \), and adapt it based on what the problem requires.
  • **Perform Calculations**: Substitute your known values into the formula and solve for the unknown.
  • **Check Your Work**: Ensure that your solution makes sense in the real world. Consider units and the answer's logical soundness.
This structured approach not only leads to the right solution but also fosters a deeper understanding of underlying principles.
Mechanics
Mechanics is the branch of physics that deals with motion and the forces that affect motion. One of its subfields is the study of energy, including gravitational potential energy. Mechanics provides a framework to understand how physical systems behave, especially concerning energy transformations. Gravitational potential energy is a key concept in mechanics, embodying the energy an object possesses due to its position relative to the Earth. When an object is elevated to a higher altitude, it gains gravitational potential energy.
  • Such energy calculations are crucial for assessing work and energy transfer in real-world scenarios, like our mountain climber problem.
  • Mechanics helps us understand how energy is conserved and transformed from one form to another.
  • It also illustrates the balance of forces acting on objects, which can result in a potential energy change.
Appreciating these principles can help students see the connections between theoretical calculations and actual physical events.

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Most popular questions from this chapter

The potential energy of two atoms in a diatomic molecule is approximated by \(U(r) = (a/r^{12}) - (b/r^6)\), where \(r\) is the spacing between atoms and \(a\) and \(b\) are positive constants. (a) Find the force \(F(r)\) on one atom as a function of \(r\). Draw two graphs: one of \(U(r)\) versus \(r\) and one of \(F(r)\) versus \(r\). (b) Find the equilibrium distance between the two atoms. Is this equilibrium stable? (c) Suppose the distance between the two atoms is equal to the equilibrium distance found in part (b). What minimum energy must be added to the molecule to \(dissociate\) it\(-\)that is, to separate the two atoms to an infinite distance apart? This is called the \(dissociation\) \(energy\) of the molecule. (d) For the molecule CO, the equilibrium distance between the carbon and oxygen atoms is 1.13 \(\times\) 10\(^{-10}\) m and the dissociation energy is 1.54 \(\times\) 10\(^{-18}\) J per molecule. Find the values of the constants \(a\) and \(b\).

A force parallel to the \(x\)-axis acts on a particle moving along the x-axis. This force produces potential energy \(U(x)\) given by \(U(x) = \alpha x^4\), where \(\alpha =\) 0.630 J/m\(^4\). What is the force (magnitude and direction) when the particle is at \(x = -0.800\) m?

A 2.50-kg mass is pushed against a horizontal spring of force constant 25.0 N/cm on a frictionless air table. The spring is attached to the tabletop, and the mass is not attached to the spring in any way. When the spring has been compressed enough to store 11.5 J of potential energy in it, the mass is suddenly released from rest. (a) Find the greatest speed the mass reaches. When does this occur? (b) What is the greatest acceleration of the mass, and when does it occur?

A small block with mass 0.0400 kg is moving in the \(xy\)-plane. The net force on the block is described by the potentialenergy function \(U(x, y) = (5.80 \, \mathrm{J/m}^2)x^2 - (3.60 \, \mathrm{J/m}^3)y^3\). What are the magnitude and direction of the acceleration of the block when it is at the point (\(x =\) 0.300 m, \(y =\) 0.600 m)?

A slingshot will shoot a 10-g pebble 22.0 m straight up. (a) How much potential energy is stored in the slingshot's rubber band? (b) With the same potential energy stored in the rubber band, how high can the slingshot shoot a 25-g pebble? (c) What physical effects did you ignore in solving this problem?
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