rt 5 of 5 Now let's try an example where we apply the work-kinetic energy relationship quantitatively to find a change in an object's speed. Let's imagine a block on a sheet of ice, with negligible friction. The block has a mass of 5.00 kg. The block is currently moving to the right at 4.00 m/s. Let's say a rope is attached to this block, and it pulls horizontally on the block, also to the right, with a force of 13.0 N. With this force applied, the block moves a distance of 3.00 m to the right. How fast is the block moving at the end of this distance? There is a lot of information in this problem, but let's think for a moment how to approach it. We can use the work-kinetic energy theorem. That is, if we can find an initial kinetic energy, and find the work, we can then solve for the final kinetic energy. Knowing the final kinetic energy, we can find the final speed. Let's first find the initial kinetic energy. What is the kinetic energy (in J) of the block when it is moving at 4.00 m/s? 40 V v Correct. This is the block's initial kinetic energy. J Next, let's calculate the work (in J) done on the block by the rope. 39 V v Correct. The work is the force times the displacement. The force on the block and the displacement are both in the same direction, so the work is positive. J Let's continue analyzing the example from the previous step. Using the initial kinetic energy and work found in the previous step, what is the final kinetic energy (in J) of the block, after it has moved the 3.00 m distance? × This is the work done on the block. How are the initial kinetic energy and the work (the two results above) related to the final kinetic energy? Use the work-kinetic energy theorem. J Now that we have the final kinetic energy, use it and the mass of the block to solve for the final speed (in m/s). × Using the result of the previous question, and the relationship between kinetic energy and speed, solve for the speed. Be careful with your algebra. m/s
rt 5 of 5 Now let's try an example where we apply the work-kinetic energy relationship quantitatively to find a change in an object's speed. Let's imagine a block on a sheet of ice, with negligible friction. The block has a mass of 5.00 kg. The block is currently moving to the right at 4.00 m/s. Let's say a rope is attached to this block, and it pulls horizontally on the block, also to the right, with a force of 13.0 N. With this force applied, the block moves a distance of 3.00 m to the right. How fast is the block moving at the end of this distance? There is a lot of information in this problem, but let's think for a moment how to approach it. We can use the work-kinetic energy theorem. That is, if we can find an initial kinetic energy, and find the work, we can then solve for the final kinetic energy. Knowing the final kinetic energy, we can find the final speed. Let's first find the initial kinetic energy. What is the kinetic energy (in J) of the block when it is moving at 4.00 m/s? 40 V v Correct. This is the block's initial kinetic energy. J Next, let's calculate the work (in J) done on the block by the rope. 39 V v Correct. The work is the force times the displacement. The force on the block and the displacement are both in the same direction, so the work is positive. J Let's continue analyzing the example from the previous step. Using the initial kinetic energy and work found in the previous step, what is the final kinetic energy (in J) of the block, after it has moved the 3.00 m distance? × This is the work done on the block. How are the initial kinetic energy and the work (the two results above) related to the final kinetic energy? Use the work-kinetic energy theorem. J Now that we have the final kinetic energy, use it and the mass of the block to solve for the final speed (in m/s). × Using the result of the previous question, and the relationship between kinetic energy and speed, solve for the speed. Be careful with your algebra. m/s
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:Part 5 of 5
Now let's try an example where we apply the work-kinetic energy relationship quantitatively to find a change in an object's speed.
Let's imagine a block on a sheet of ice, with negligible friction. The block has a mass of 5.00 kg. The block is currently moving to the right at 4.00 m/s. Let's say a rope is attached to this block, and it pulls horizontally on the
block, also to the right, with a force of 13.0 N. With this force applied, the block moves a distance of 3.00 m to the right. How fast is the block moving at the end of this distance?
There is a lot of information in this problem, but let's think for a moment how to approach it.
We can use the work-kinetic energy theorem. That is, if we can find an initial kinetic energy, and find the work, we can then solve for the final kinetic energy. Knowing the final kinetic energy, we can find the final speed.
Let's first find the initial kinetic energy. What is the kinetic energy (in J) of the block when it is moving at 4.00 m/s?
40
V v Correct. This is the block's initial kinetic energy. J
Next, let's calculate the work (in J) done on the block by the rope.
39
V Correct. The work is the force times the displacement. The force on the block and the displacement are both in the same direction, so the work is positive. J
Let's continue analyzing the example from the previous step.
Using the initial kinetic energy and work found in the previous step, what is the final kinetic energy (in J) of the block, after it has moved the 3.00 m distance?
x This is the work done on the block. How are the initial kinetic energy and the work (the two results above) related to the final kinetic energy? Use the work-kinetic energy theorem. J
Now that we have the final kinetic energy, use it and the mass of the block to solve for the final speed (in m/s).
x Using the result of the previous question, and the relationship between kinetic energy and speed, solve for the speed. Be careful with your algebra. m/s
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