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$\bullet$ A 12.0 g plastic ball is dropped from a height of 2.50 $\mathrm{m}$ andis moving at 3.20 $\mathrm{m} / \mathrm{s}$ just before it hits the floor. How muchmechanical energy was lost during the ball's fall?

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0.233 .1

03:26

Surjit Tewari

Physics 101 Mechanics

Chapter 7

Work and Energ

Physics Basics

Applying Newton's Laws

Kinetic Energy

Potential Energy

Energy Conservation

Cornell University

Simon Fraser University

Hope College

University of Winnipeg

Lectures

03:47

In physics, the kinetic energy of an object is the energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest. The kinetic energy of a rotating object is the sum of the kinetic energies of the object's parts.

04:05

In physics, a conservative force is a force that is path-independent, meaning that the total work done along any path in the field is the same. In other words, the work is independent of the path taken. The only force considered in classical physics to be conservative is gravitation.

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in this question. We have a 12 grand plastic ball dropped from a height of 2.5 m and it is moving 3.2 m per second just before it hits the floor. Were asked to calculate how much mechanical energy is lost during the balls fall. So basically, what this question is asking for is three change in energy, the change in mechanical energy, um, between the initial height and then just before it hits the floor. So we're gonna be calculating Delta E the change in energy. So that's going to be e to sow e final minus e initial e one. Now, the thing about mechanical energy is that it is usually composed of both kinetic energy and, um, potential energy in this case, gravitational potential energy. So each of these terms E two and E one has a potential to include both kinetic energy and gravitational energy. But let's think of our actual scenario here. So, just before the ball hits the ground, um, all of the gravitational potential energy has been converted into kinetic energy. So in the final state, the only thing that is present is kinetic energy, so we can write e two as one half, um, B two squared re to being the final speed. And then in the initial stage just before the ball is dropped We have no kinetic energy because the ball is not moving and we Onley have gravitational potential. So for the initial state, we're going to write MGI um h one Ah, we all need to include the gravitational potential in that in that term. So we just go ahead and plug in all of our given. So the mass. Of course, we need to convert that into kilograms. So that's going to be 0.12 kg. The velocity, 3.2 m per second squared and then we do the mass again times g times the H that we were given 2.5 eaters. So the first term when we calculate that about 0.2943 jewels the second term 0.6 144 jewels. So when we do that subtraction, we get about 0.233 jaws. Actually, um, I've written thes backwards, so it would be 0.6144 jewels minus 0.294 three jewels. So that actually gives us negative 0.2 33 jewels. And the reason why that negative is significant is that means that we have lost some energy, right? So something here has taken energy out off the ball. Um, presumably, since it's falling through the air, it's probably a drag force. Um, that's doing some ah, work on the ball and taking some energy out s So this is your final answer here.

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