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A \(2.50 \mathrm{~kg}\) beverage can is thrown directly downward from a height of \(4.00 \mathrm{~m}\), with an initial speed of \(3.00 \mathrm{~m} / \mathrm{s}\). The air drag on the can is negligible. What is the kinetic energy of the can (a) as it reaches the ground at the end of its fall and (b) when it is halfway to the ground? What are (c) the kinetic energy of the can and (d) the gravitational potential energy of the can-Earth system \(0.200 \mathrm{~s}\) before the can reaches the ground? For the latter, take the reference point \(y=0\) to be at the ground.

Short Answer

Expert verified
(a) KE at ground: 109.35 J, (b) KE halfway: 60.30 J, (c) KE 0.200s before: 86.95 J, (d) PE 0.200s before: 39.95 J.

Step by step solution

01

Identify the Initial Parameters

Given: - Mass of the can, \( m = 2.50 \, \text{kg} \) - Initial height, \( h = 4.00 \, \text{m} \) - Initial speed, \( v_i = 3.00 \, \text{m/s} \) - Gravitational acceleration, \( g = 9.81 \, \text{m/s}^2 \) The air drag is negligible, so we have a freely falling object and conservation of mechanical energy applies.
02

Calculate Final Speed at Ground

Using the kinematic equation for final velocity \( v_f^2 = v_i^2 + 2gh \):\[ v_f^2 = (3.00)^2 + 2 \times 9.81 \times 4.00 \] \[ v_f^2 = 9 + 78.48 = 87.48 \]\[ v_f = \sqrt{87.48} \approx 9.35 \, \text{m/s} \]
03

Calculate Kinetic Energy at the Ground

Kinetic Energy (KE) is given by: \[ KE = \frac{1}{2} mv^2 \]Substituting the final speed:\[ KE = \frac{1}{2} \times 2.50 \times (9.35)^2 \]\[ KE = \frac{1}{2} \times 2.50 \times 87.48 \]\[ KE = 109.35 \, \text{J} \]
04

Calculate Speed and Energy Halfway Down

Halfway to the ground, the height is \( 2.00 \, \text{m} \). Use kinematic equations:\[ v_{half}^2 = v_i^2 + 2gh_{half} \] \[ v_{half}^2 = (3.00)^2 + 2 \times 9.81 \times 2.00 \]\[ v_{half}^2 = 9 + 39.24 = 48.24 \]\[ v_{half} = \sqrt{48.24} \approx 6.95 \, \text{m/s} \]Now, calculate KE:\[ KE = \frac{1}{2} \times 2.50 \times (6.95)^2 \]\[ KE = \frac{1}{2} \times 2.50 \times 48.24 \]\[ KE = 60.30 \, \text{J} \]
05

Velocity & KE at 0.200s Before Hitting the Ground

To find the velocity \( 0.200 \, \text{s} \) before it reaches the ground, use: \[ v = v_i + gt \]Here, \( t \) is the total time minus \( 0.200 \, \text{s} \). We first calculate total fall time using \( y = v_i t + \frac{1}{2}gt^2 \):\[ 4.00 = 3.00t + \frac{1}{2} \times 9.81 \times t^2 \]Solving this quadratic equation:\( t\approx0.744 \, \text{s} \)Thus at \(0.200 \, \text{s}\) before it hits the ground, \(t = 0.544 \) s.Calculate velocity:\[ v = v_i + gt \approx 3.00 + 9.81 \times 0.544 \approx 8.34 \, \text{m/s} \]Calculate KE:\[ KE = \frac{1}{2} \times 2.50 \times (8.34)^2 \]\[ KE = 86.95 \, \text{J} \]
06

Calculate Gravitational Potential Energy 0.200s Before Ground

Using the height \( y \) at \( 0.544 \mathrm{~s} \) before hitting the ground:\[ y = 3.00 \times 0.544 + \frac{1}{2} \times 9.81 \times (0.544)^2 \]\[ y \approx 1.63 \, \text{m} \]Potential Energy (PE) at this point:\[ PE = mgy \]\[ PE = 2.50 \times 9.81 \times 1.63 \]\[ PE = 39.95 \, \text{J} \]

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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 an object possesses due to its position in a gravitational field. In this context, we consider it for the can as it falls from a height.
When the can is at its initial height of 4.00 meters, it has a certain amount of GPE, calculated using the formula:
  • \[ PE = mgh \]
where:
  • \( m \) is the mass of the can (2.50 kg)
  • \( g \) is the acceleration due to gravity (9.81 m/s²)
  • \( h \) is the height (4.00 m initially)
As the can falls, its height decreases and so does its GPE. The key idea here is that the decrease in potential energy will convert into kinetic energy (KE), as there is no air resistance to consider. This is further explained by the concept of conservation of mechanical energy.
Conservation of Mechanical Energy
In the absence of non-conservative forces like air resistance, the principle of conservation of mechanical energy states that the total mechanical energy in a closed system remains constant.
For the falling can:
  • Total mechanical energy = Kinetic energy (KE) + Gravitational potential energy (PE)
During the fall, gravitational potential energy decreases because the height decreases.
At the same time, kinetic energy increases because the can accelerates.
The sum of PE and KE at any point during its fall will always equal the initial total mechanical energy at the starting height.
By this principle, when the can reaches the ground, all the initial potential energy has transformed into kinetic energy.
This transformation allows us to calculate unknowns if some values are given, as seen in the problem.
Kinematics
Kinematics is a branch of mechanics that deals with the motion of objects without considering the forces causing the motion.
In the scenario of a can falling under gravity, several kinematic equations help determine characteristics like velocity and displacement.
  • One essential equation is:\[ v_f^2 = v_i^2 + 2gh \]where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, and \( h \) is the height.
This equation is crucial in problems like finding how fast the can moves at various points.
Another critical kinematic concept is the time of fall, calculated using:
  • \[ y = v_i t + \frac{1}{2}gt^2 \]
This finds the height at any time during a fall, allowing us to track the can’s position and speed accurately as well as calculating PE and KE at specific points.
Free Fall
Free fall describes the motion of an object under the influence of gravity alone, without any air resistance.
This is a perfect simplification for calculations in physics as it involves only gravitational acceleration.
  • The gravitational force pulls the can downwards with a constant acceleration of 9.81 m/s² (on Earth's surface).
In this problem, when the can is released, it accelerates downward at this constant rate.
No other forces are acting upon it except gravity, meaning the principles of free fall and conservation of energy can be directly applied. Understanding this concept simplifies our task of predicting its speed and energy at different points till it reaches the ground.
The initial push simply gives it a head start, but from the moment it is released, its acceleration is constant thanks to gravity.

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