# Work and Energy Theorem and Kinetic Energy

• rmarkatos
In summary, the problem involves a 5.0x10^4kg space probe traveling at a speed of 11,000m/s through deep space. Retrorockets are fired to reduce its speed by generating a force of 4.0x10^5N over a distance of 2500km. Using the equation W=Fd=1/2m(vf^2)-1/2m(vi^2), there seems to be a discrepancy in finding the final speed of the probe, as the correct answer of 9,000m/s does not match the solution using this equation. Further clarification and steps may be needed to correctly solve the problem.
rmarkatos
A 5.0x10^4kg space probe is traveling at a speed of 11,000m/s through deep space. Retrorockets are fired along the line of motion to reduce the probe's speed. The retrorockets generate a force of 4.0x10^5N over a distance of 2500km. What is the final speed of the probe.

For this problem i know exactly what to do but i know i am getting the wrong answer because I have odd answer selections.

I used W=Fd=1/2m(vf^2)-1/2m(vi^2). I know this is correct because when i plug in the correct answer for vf=9,000 m/s it equals the other side of the equation so i know i am right but for some reason in the process of solving i am not sure what i am doing wrong because i keep getting the wrong answer for vf. It is kind of strange because my units in every part of solving the equation come out correctly.

Any suggestions?

Are you manipulating the equality correctly? Perhaps if you were to show your steps we may be able to comment further.

Yes, sorry about that i should of.

First i converted km to meters because we work in the mks system

(4x10^5N)(2.5X10^6m)=w
=1x10^12J

1x10^12J=1/2(5x10^4kg)(vf^2)-1/2(5x10^4kg)(11,000m/s)^2

1x10^12J=1/2(5x10^4kg)(vf^2)-3.025J

from here i tried different ways to solve the equation but each way i still get the wrong answer. I added the 3.025J to both sides and then multiplied the 1/2(5x10^4). Then i divided the 4.025J by the 25,000kg and then solved for vf by taking the square root of both sides but for some reason my answer does not really come out relativlely close 9000m/s

rmarkatos said:
1x10^12J=1/2(5x10^4kg)(vf^2)-3.025J
Do you not think this number looks a little small?

Oh sorry a bit of typo there. It is supposed to be-3.025x10^12. Sorry i forgot to type that in there but when i was doing the problem i did have it correctly written.

## What is the Work and Energy Theorem?

The Work and Energy Theorem is a fundamental principle in physics that states that the work done on an object is equal to the change in its kinetic energy. In other words, the work-energy theorem is a mathematical representation of the relationship between work and energy.

## What is Kinetic Energy?

Kinetic energy is the energy that an object possesses due to its motion. It is a scalar quantity and is dependent on an object's mass and speed. The formula for calculating kinetic energy is KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

## How is the Work and Energy Theorem related to Kinetic Energy?

The Work and Energy Theorem states that the work done on an object is equal to the change in its kinetic energy. This means that when work is done on an object, its kinetic energy will either increase or decrease depending on the direction of the work. In other words, the work done on an object is directly related to its change in kinetic energy.

## What are some real-world applications of the Work and Energy Theorem and Kinetic Energy?

The Work and Energy Theorem and Kinetic Energy are important concepts in many fields, such as engineering, mechanics, and even sports. For example, engineers use these principles to design efficient machines and structures, while athletes use their understanding of kinetic energy to improve their performance in sports like track and field or gymnastics.

## How does the Work and Energy Theorem and Kinetic Energy relate to conservation of energy?

The Work and Energy Theorem and Kinetic Energy are closely related to the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred or transformed. In the case of the work-energy theorem, the work done on an object is converted into kinetic energy, which means that the total amount of energy in the system remains constant. This principle is essential in understanding the behavior of many physical systems.

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