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Conservation of Energy and Projectile Motion Recall the projectile motion problem from chapter two.  The picture above showed the maximum height and time to reach that height from a cannon shot at a 30.0 degree angle with a speed of 150 m/s. Solving for the maximum height required the use of three equations in chapter two: Vy equation, Vf equation, and the displacement equation. The conservation equation allows us to find the maximum height with one equation. (Once we know the y-velocity). This is where it is important to remember that the conservation equation only deals with vertical velocities. So, you cannot plug 150 m/s into the equation. From chapter two, the Vy component was found to be 75 m/s. Therefore, the conservation equation would become, Again, since the vertical speed at the top of the trajectory is zero, Vf is set at zero. Also, we assume ho is zero, and the masses cancel. Solving for hf indeed gives the answer of 287 meters! There are more examples that utilize the conservation of energy equation covered in the questions archive page under chapter four. I just wanted to point out previous problems to show the correlation between the two methods. The main thing to remember when tackling different problems is to work from left to right in the conservation equation. The Pile Driver Consider the picture below.  This is my verson of the pile driver. You have a wench that runs a rope over a pulley and is attached to a 1,000-kg object. (a little over a ton). When the machine releases the driver, it crashes into the spike, thrusting it into the ground. I want to talk about this machine in terms of energy. while the driver is suspended above the spike, it has potential energy with respect to the spike's head. Let's say it is 10 meters, about 33 feet. knowing the mass, the potential energy is (1000)(9.8)(10) which equals 98,000 Joules.  After the driver is realeased, the potential is converted to kinetic until all of the potential becomes kinetic right before the instant it hits the spike. Again, we know that the kinetic energy at that point is equal to the initial potential energy.  After impact, all of the kinetic energy is converted to work on the spike. The work drives the spike into the ground a certain distance with a certain force.  So, we can determine how much force the driver used if we measured the distance the spiken was driven into the ground. Let's say the distance was 10 cm which is .1 meters.  The work equation would be, Which gives a force of 980,000 Newtons. Dividing by 4.448 to convert to pounds gives a force of 220,000 pounds! The point of this example was to again illustrate how energy is transfered from one form to another, and how potential and kinetic energy relates to work. There are some very interesting aspects to consider from this problem. For example, if I wanted the driver to do twice the work on the spike, how high would it need to be raised? Since potential is directly related to work, you would just have to double the height. However, how high would you have to raise the driver so that it would hit the spike with twice the speed? If you said twice as high, you are wrong. Potential energy is directly related to the speed squared because in this example, This is the same relationship that work has with kinetic energy. Therefore, you would have to raise the driver to four times the height to achieve just twice the speed. This would in turn result in four times the work. Perhaps the most intriguing aspect involves this question: Which situation would yield the greatest force between driver and spike, the spike being driven in a large distance or a short distance? Most people would automatically say the large distance because it took a large force to do that. This is another instance where physics seems to counter common sense. It turns out that if the spike is driven into the ground a small distance, the driver would hit it with a larger force compared to the spike being driven in a large distance! The reason has to do with the fact you only have a certain amount of work available. Remember these concept equations?  When the distance is small, the force is large so that the product will equal the available amount of work. It actually helps to think about it in terms of soil. If the ground is soft, it stands to reason that the spike will be driven in a larger distance without much effort from the driver. On the other hand, if the ground is very dry and hard, the spike will not go in as far, and the driver will be slamming into this rigidly set spike. Which one would you rather hit with your hand, the spike in the soft soil or the spike in the rigid soil? The spike in the rigid soil would hurt more why? Because you are hitting it with a larger force. This also applies to the situation of stopping a car. If you stop a car over a small distance, the force is greater. Same with the spike and driver. You can think of it as not the driver hiting the spike, but the spike stopping the driver. So, if the spike stops the driver over a short distance, it will need a larger force to do so and vice versa. Hopefully, these perspectives shed some light on this concept. |