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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) The final kinetic energy is calculated using the final velocity. (b) Kinetic energy halfway involves halfway velocity. (c) Use velocity 0.2s before impact for kinetic energy, (d) and height for potential energy 0.2s before impact.

Step by step solution

01

Identify Known Parameters

First, note that the mass of the can is \( m = 2.50 \, \text{kg} \), the initial height is \( h_0 = 4.00 \, \text{m} \), the initial speed is \( v_0 = 3.00 \, \text{m/s} \), and acceleration due to gravity is \( g = 9.81 \, \text{m/s}^2 \). We'll also consider \( y = 0 \) at the ground.
02

Calculate Final Velocity

Use the equation of motion for velocity: \[ v^2 = v_0^2 + 2gh \]Substitute \( v_0 = 3.00 \, \text{m/s} \), \( g = 9.81 \, \text{m/s}^2 \), and \( h = 4.00 \, \text{m} \) to find the final velocity \( v \) when the can hits the ground.
03

Compute Final Kinetic Energy

The kinetic energy as the can hits the ground can be found using the formula \[ KE = \frac{1}{2}mv^2 \]Substitute the final velocity \( v \) calculated in the previous step and \( m = 2.50 \, \text{kg} \) to find the kinetic energy at the ground.
04

Determine Velocity at Halfway Point

Calculate the velocity at halfway to the ground using the equation: \[ v^2 = v_0^2 + 2g\left(\frac{h}{2}\right) \]Substitute the known values to solve for this halfway velocity.
05

Compute Kinetic Energy at Halfway Point

Again use the kinetic energy formula \[ KE = \frac{1}{2}mv^2 \]with the velocity from the previous step to find the kinetic energy halfway to the ground.
06

Find Position 0.2s Before Impact

Calculate the position above ground 0.2 seconds before impact using: \[ y = v_f t - \frac{1}{2}gt^2 \]where \( t = 0.2 \, \text{s} \) and \( v_f \) is the calculated final velocity.
07

Calculate Velocity 0.2s Before Impact

Use the velocity equation to find it when \( t = 0.2 \, \text{s} \) before impact: \[ v = v_f - gt \].
08

Compute Kinetic Energy 0.2s Before Impact

Substitute into \[ KE = \frac{1}{2}mv^2 \]with the velocity found 0.2 seconds before impact.
09

Compute Gravitational Potential Energy

Use the potential energy formula \[ PE = mgy \]with \( y \) being the height 0.2 seconds before hitting the ground.

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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 is a form of energy that an object possesses due to its position in a gravitational field. It is especially important when considering objects in free fall, as it helps us understand how energy is transferred from one type to another as the object moves. For the beverage can exercise, we consider gravity's effect from a height of 4.00 meters.
To calculate gravitational potential energy (PE), the formula is used:\[ PE = mgh \]where:
  • \( m \) is the mass of the object, in this case, the can (2.50 kg).
  • \( g \) is the acceleration due to gravity, which averages 9.81 m/s² on Earth.
  • \( h \) is the height of the object above the reference point (usually the ground).
As the can falls, its potential energy decreases because \( h \) becomes smaller, converting into kinetic energy. This concept is fundamental in energy conservation, as the total energy (potential + kinetic) remains constant in a closed system with negligible air resistance.
Equations of Motion
Equations of motion play a critical role in predicting the movement of objects under various forces, which includes the actions of gravity. These equations allow us to solve for unknown variables when an object is in motion, particularly in scenarios like the free fall of our beverage can.
There are a few key equations that help analyze the motion:
  • For velocity: \[ v^2 = v_0^2 + 2gh \]
  • For position: \[ y = v_f t - \frac{1}{2}gt^2 \]
Each equation has specific variables:
  • \( v \) is the final velocity, the speed of the can after falling a certain distance.
  • \( v_0 \) is the initial velocity, which in this exercise is the starting speed of the can (3.00 m/s).
  • \( g \) is the gravitational acceleration (9.81 m/s²).
  • \( t \) is the time, particularly used in calculating how far an object travels in a given timespan.
These formulas allow us to calculate velocities at different points of the fall and understand how energy transfers from potential to kinetic as the can accelerates towards the ground.
Free Fall Motion
Free fall motion occurs when the only force acting upon an object is gravity, causing it to accelerate downward. In this exercise, as the can is thrown downward, we assume that no other forces, like air resistance, affect it besides gravity.
Key characteristics of free fall include:
  • Constant Acceleration: The object accelerates due to gravity at 9.81 m/s².
  • Initial Speed: In this problem, the can starts with an initial downward velocity of 3.00 m/s.
Kinetic energy changes significantly throughout the free fall. Initially, the kinetic energy is calculated using:\[ KE = \frac{1}{2}mv^2 \]where \( m = 2.50 \text{ kg} \) is the mass and \( v \) is the velocity at a given point.
As the can descends, its velocity increases due to the constant gravitational pull, leading to a rise in its kinetic energy. Over different stages of the fall, kinetic and potential energies transform but within the system, the total mechanical energy remains consistent. This offers an excellent example of conservation of energy and the influence gravity has on moving objects.

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