# 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 along the line of motion to reduce the probe's speed, 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), the correct answer for vf=9,000m/s is obtained, but the units and algebraic steps are causing confusion and leading to incorrect answers. The issue may be resolved by considering the negative work done by the rockets, as it opposes the motion of the
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.
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)-3.025x10^12J

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

Any suggestions?

The work done by the rockets is negative since their force opposes the motion.

i am just having a problem solving it algebraically correctly for some reason.

Because your work should be negative. The final kinetic energy - the initial kinetic energy should be negative, yet you have set it to a positive value, meaning v final is going to be greater than v initial. Try making that LHS negative, and it should work.

## 1. What is the Work and Energy Theorem?

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 the amount of work done on an object will result in a change in its speed or velocity.

## 2. How is kinetic energy related to the Work and Energy Theorem?

Kinetic energy is the energy an object possesses due to its motion. In the context of the Work and Energy Theorem, the change in an object's kinetic energy is directly proportional to the work done on the object. This means that as more work is done on an object, its kinetic energy will increase.

## 3. How is the Work and Energy Theorem applied in real-world situations?

The Work and Energy Theorem is used to understand and analyze the movement of objects in various scenarios, such as collisions, falling objects, and moving vehicles. It allows us to calculate the work done on an object and its resulting change in kinetic energy.

## 4. Is the Work and Energy Theorem always applicable?

While the Work and Energy Theorem is a fundamental concept in physics, it may not always be applicable in certain situations. For example, if there are non-conservative forces (such as friction) acting on the object, the work done by these forces must also be taken into account.

## 5. How does the Work and Energy Theorem relate to the law of conservation of energy?

The Work and Energy Theorem is essentially a manifestation of the law of conservation of energy. This law states that energy cannot be created or destroyed, only transferred or transformed. In the context of the Work and Energy Theorem, the work done on an object results in a change in its kinetic energy, but the total energy of the system remains constant.

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