Change in the Kinetic Energy of a System

In summary, the conversation discusses finding the change in kinetic energy of a system consisting of a pulley system with a wooden block and a 100g mass on a flat surface. The force of friction and individual speeds at every 0.1 of a second are calculated using a ticker timer, and the distance traveled is 45cm. The equation for kinetic energy is mentioned and the need to prove that the change in kinetic energy is equal to the work done in the system. The variables and velocities to use in the calculation are also considered, as well as the role of friction in kinetic energy. It is mentioned that a diagram would be helpful in fully understanding the situation.
  • #1
drewcila
1
0
OP warned about not using the homework template
The question is to find the change in kinetic energy of a system. the system in use is a pully system with a wooden block on a flat surface attached to a 100g mass. I have calculated force of friction between the block and desk, and using a ticker timer, I have individual speeds at every 0.1 of a second. Additionally, the distance it travels is 45cm. We have to prove that the change in Kinetic energy is = work done in the system, so I can't use work as my answer.

I know that the equation for kinetic energy is Ek= 1/2mv2
Change in Kinetic energy is represented by the final Ek - initial Ek

I was thinking something like Ekf = 1/2(0.100)v22
then subtract Eki= 1/2(0.100)v12
I do not know what variables to use in this situation. do I use the mass of the block or the 100g mass apply force? and which velocity should I use? the speed at 0.1 seconds and 0.6 seconds? Does friction play into kinetic energy?
 
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  • #2
drewcila said:
We have to prove that the change in Kinetic energy is = work done in the system, so I can't use work as my answer.
I am not sure what this means. It looks like you are asked to prove the work-energy theorem, ##\Delta K = W_{net}## so you do need to calculate work. Also, is your wooden block pulled by a hanging mass through a string over a pulley? If so, the change in kinetic energy is the difference in kinetic energy between two points in space. Your system has two moving masses, so you need to calculate the kinetic energy of each. Note that they are moving at the same speed at all times. So the left side of the equation is ##\Delta K=\Delta K_{wood}+\Delta K_{mass}##. You can calculate this and get a number.

The right side of the equation is the total work done by all the forces acting on the system. You need to figure out what forces do how much work on what parts of the system, calculate the numbers and add them all up. This number you get for the right side of the equation should be equal to the first number you got for the left side of the equation, ##\Delta K##.
 
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  • #3
A diagram would be helpful
 

1. How is kinetic energy defined?

Kinetic energy is the energy an object possesses due to its motion. It is defined as one half of the object's mass multiplied by the square of its velocity (KE = 1/2 * m * v^2).

2. What factors affect the change in kinetic energy of a system?

The change in kinetic energy of a system is affected by the mass and velocity of the objects involved. An increase in mass or velocity will result in a greater change in kinetic energy.

3. How is the change in kinetic energy of a system calculated?

The change in kinetic energy of a system can be calculated by subtracting the initial kinetic energy from the final kinetic energy.

4. Can the change in kinetic energy of a system be negative?

Yes, the change in kinetic energy of a system can be negative if the final kinetic energy is less than the initial kinetic energy. This can happen if the object experiences a decrease in velocity or if the direction of its motion changes.

5. How does friction impact the change in kinetic energy of a system?

Friction can cause a decrease in the change in kinetic energy of a system by converting some of the kinetic energy into heat. This results in a slower change in the object's velocity and a smaller overall change in kinetic energy.

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