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Chapter 5: Problem 19

A \(0.20-\mathrm{kg}\) stone is held \(1.3 \mathrm{~m}\) above the top edge of a water well and then dropped into it. The well has a depth of \(5.0 \mathrm{~m}\). Taking \(y=0\) at the top edge of the well, what is the gravitational potential energy of the stone-Earth system (a) before the stone is released and (b) when it reaches the bottom of the well. (c) What is the change in gravitational potential energy of the system from release to reaching the bottom of the well?

Short Answer

Expert verified
The gravitational potential energy of the stone-Earth system (a) before the stone is released is 2.548J and (b) when it reaches the bottom of the well is 12.348J. (c) The change in gravitational potential energy of the system from release to reaching the bottom of the well is 9.8J.

Step by step solution

01

Calculate initial gravitational potential energy

Calculation of potential energy at the released position is done using the formula, \(PE_{initial}=m*g*h\). Given, mass, \(m= 0.20kg\), gravitational acceleration, \(g = 9.8m/s^2\) and \(h=1.3m\). substituting these values into the formula, the initial potential energy is calculated as: \( PE_{initial} = 0.20kg * 9.8m/s^2 * 1.3m = 2.548J\).
02

Calculate gravitational potential energy at the bottom of the well

To calculate the potential energy at the bottom of the well, consider the total height from the top of the well to the bottom, which is \(1.3m (initial height) + 5.0m (depth of the well) = 6.3m\). Then use the formula \(PE_{final}=m*g*h\), after substitution we have: \(PE_{final} = 0.20kg * 9.8m/s^2 *6.3m = 12.348J\)
03

Calculate change in gravitational potential energy

The change in potential energy is calculated by subtracting \(PE_{initial}\) from \(PE_{final}\). So, \(\Delta PE = PE_{final} - PE_{initial} = 12.348J - 2.548J = 9.8J\)

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Key Concepts

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

Energy Conservation
Energy conservation is a powerful principle in physics stating that energy in a closed system remains constant. In the context of gravitational potential energy, this means that as the stone falls into the well, the energy doesn't disappear; it transforms. The initial gravitational potential energy of the stone is converted into kinetic energy as it drops.
  • Gravitational potential energy is highest when the object is at its initial point before being released.
  • Kinetic energy increases as the stone falls, demonstrating the conservation of energy.
  • When the stone reaches the bottom of the well, its kinetic energy is at its maximum, assuming negligible air resistance.
Understanding energy conservation helps us predict how energies transform and balance, offering insight into how systems behave both theoretically and in practical applications.
Physics Problem Solving
Approaching physics problems, particularly involving gravitational potential energy, requires a structured method. This helps in accurately calculating energies and understanding the processes involved.
  • Identify the known values and what is being asked in the problem (such as mass, gravitational acceleration, heights, and energy changes).
  • Write down the relevant equations such as those for gravitational potential energy: \(PE = m \cdot g \cdot h\).
  • Substitute known values carefully into the equations to compute the unknowns.
  • Verify the units utilized, ensuring consistency, since incorrect units can lead to errors in calculations.
This systematic approach allows students to solve problems accurately and is a great practice for mastering physics concepts.
Kinematics
Kinematics is the branch of physics that describes the motion of objects without considering the forces causing it. In this scenario, it helps us understand the motion of the stone as it's dropped into the well.
  • The stone starts from rest, implying its initial velocity is zero.
  • As the stone falls, it accelerates downward under the influence of gravity, reaching higher speeds as time progresses.
  • The equations of motion can predict the speed of the stone at various depths of the well until it reaches the bottom.
Although the calculations of potential energies are central, recognizing how the stone's motion is described in kinematics complements the understanding of its energy transformations.
Gravity
Gravity is a fundamental force that plays a crucial role in many physical processes, including how energy is stored and converted in objects. In this problem, gravity is responsible for the stone's motion and potential energy.
  • The gravitational force on any object is calculated by multiplying the mass of the object by the gravitational acceleration (on Earth, approximately \(9.8 \text{ m/s}^2\)).
  • This force causes objects to accelerate towards the center of the Earth, influencing both their potential and kinetic energies.
  • Gravitational potential energy is directly tied to the height of the object, emphasizing gravity's role in energy calculations.
Understanding gravity's impact is essential not only in solving such physics problems but also in wider applications ranging from simple mechanics to astrophysics.

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Most popular questions from this chapter

QIC The masses of the javelin, discus, and shot are \(0.80 \mathrm{~kg}, 2.0 \mathrm{~kg}\), and \(7.2 \mathrm{~kg}\), respectively, and record throws in the corresponding track events are about \(98 \mathrm{~m}, 74 \mathrm{~m}\), and \(23 \mathrm{~m}\), respectively. Neglecting air resistance, (a) calculate the minimum initial kinetic energies that would produce these throws, and (b) estimate the average force exerted on each object during the throw, assuming the force acts over a distance of \(2.0 \mathrm{~m}\). (c) Do your results suggest that air resistance is an important factor?

\(Q \mid C\) (a) A child slides down a water slide at an amusement park from an initial height \(h\). The slide can be considered frictionless because of the water flowing down it. Can the equation for conservation of mechanical energy be used on the child? (b) Is the mass of the child a factor in determining his speed at the bottom of the slide? (c) The child drops straight down rather than following the curved ramp of the slide. In which case will he be traveling faster at ground level? (d) If friction is present, how would the conservation-ofenergy equation be modified? (e) Find the maximum speed of the child when the slide is frictionless if the initial height of the slide is \(12.0 \mathrm{~m}\).

A child's pogo stick (Fig. P5.81) stores energy in a spring \((k=\) \(\left.2.50 \times 10^{4} \mathrm{~N} / \mathrm{m}\right)\). At position (A) \(\left(x_{1}=-0.100 \mathrm{~m}\right)\), the spring compression is a maximum and the child is momentarily at rest. At position (B) \((x=0)\), the spring is relaxed and the child is moving upward. At position (C), the child is again momentarily at rest at the top of the jump. Assuming that Figure p \(5.81\) the combined mass of child and pogo stick is \(25.0 \mathrm{~kg}\), (a) calculate the total energy of the system if both potential energies are zero at \(x=0\), (b) determine \(x_{2}\), (c) calculate the speed of the child at \(x=0\), (d) determine the value of \(x\) for which the kinetic energy of the system is a maximum, and (c) obtain the child's maximum upward speed.

QIC A \(0.250-\mathrm{kg}\) block along a horizontal track has a speed of \(1.50 \mathrm{~m} / \mathrm{s}\) immediately before colliding with a light spring of force constant \(4.60 \mathrm{~N} / \mathrm{m}\) located at the end of the track. (a) What is the spring's maximum compression if the track is frictionless? (b) If the track is not frictionless, would the spring's maximum compression be greater than, less than, or equal to the value obtained in part (a)?

In the dangerous "sport" of bungee jumping, a daring student jumps from a hot- air balloon with a specially designed elastic cord attached to his waist. The unstretched length of the cord is \(25.0 \mathrm{~m}\), the student weighs \(700 \mathrm{~N}\), and the balloon is \(36.0 \mathrm{~m}\) above the surface of a river below. Calculate the required force constant of the cord if the student is to stop safely \(4.00 \mathrm{~m}\) above the river.
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