A 2.50 kg beverage can is thrown directly downward from a...

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.

Key Concepts

Conservation of Mechanical Energy

The principle of conservation of mechanical energy states that in the absence of non-conservative forces (such as significant air resistance), the total mechanical energy of a system (the sum of kinetic and potential energy) remains constant. This concept allows one to relate the kinetic and potential energies at different points in an object's motion, especially in free-fall scenarios.

Kinematics of Free Fall

Kinematics of free fall involves analyzing the motion of objects under the influence of gravity alone, where the acceleration is constant. It includes calculating parameters such as displacement, velocity, and time using equations of motion. This concept provides the necessary tools to determine how the speed of a falling object changes with time or distance, which is essential for energy calculations in these contexts.

Gravitational Potential Energy

Gravitational potential energy is the energy stored in an object as a result of its position in a gravitational field. It is typically calculated using the formula mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height above a reference point. This concept is crucial in understanding energy conversion between potential and kinetic forms.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. It is calculated by the expression (1/2)mv², where m is the mass of the object and v is its velocity. This concept is fundamental in mechanics for determining how energy is distributed in moving systems and for analyzing energy changes as objects accelerate or decelerate.

Transcript

00:01 Okay, part a.

00:04 Applying conservation of energy, initial kinetic energy plus initial potential energy is equal to final kinetic energy plus final potential energy.

00:22 And this equation can be written as ki plus ui minus uf is equal to kf and putting the values we have kf is equal to final kinetic energy is equal to half m is 2 .5 multiply by v square v is 3 plus 2 .5 and u i minus uf is equal to m gh therefore we have m is 2 .5 and g is 9 .8 and h is equal to 4.

01:21 Therefore, kinetic energy final is equal to 109 .25 jules.

01:32 So kinetic energy final is equal to 109 .25 jules.

01:39 Part b.

01:41 Now let's solve part b.

01:44 Again, applying conservation of energy, we have a kinetic energy initial k i plus u i is equal to k h w plus u h w plus u h w and here kh w is the kinetic energy of the can when it's halfway to the ground so kh w is the kinetic energy of the can when it's halfway to the ground so khw is the kinetic energy of the can when it's halfway to the ground similarly uh w is the potential energy of the kn when it's when it is halfway to the ground this can be written as k i plus ui minus u hw hw halfway is equal to kinetic energy 8 halfway so we have k halfway kinetic energy energy halfway is equal to 1 divided by 2 multiply by 2 .5 multiply by 3 square plus now u i minus u halfway is the change in potential energy and in this case it is m g h halfway so it is m g h halfway and m is 2 .5 multiply by g is 9 .8 and halfway is 4 divided by 2 that's equal to 4 divided by 2 and therefore kinetic energy halfway is equal to 60 .25 joules so kinetic energy of the cane halfway to the ground is 60 .25 jules.

04:19 Part c now we will solve part c well here kinetic energy final kf is equal to half m vf square where vf is the final velocity and we have final kinetic energy it's equal to 109 .25 and it's equal to 1 divided by 2 multiplied by m is 2 .5 multiplied by vf square and therefore we have vf is equal to square root of 2 multiply by 109 .25 divided by 2 .5 divided by 2 .5 divided by 2 .5 and it's equal to 9 .35 meter per second 9 .35 meter per second and we know that vf is equal to vi plus at and putting the values we have 9 .35 is equal to 3 plus 80...

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