# Apply the law of conservation of energy to an object

• annabelx4
In summary, the law of conservation of energy states that in a closed system without nonconservative forces, the total mechanical energy is conserved. This applies to objects launched upward in the gravitational field of the earth, where the energy transformations involve kinetic and potential energy. The equation K_i + U_1 + W_other = K_2 + U_2 represents this conservation of energy, where "i" represents the initial moment and "f" represents the final moment. In terms of solving for the object's speed at a given height, the equation (1/2)mv^2 + 0 + 0 = (1/2)mv^2 + mg(v^2/4g) can be used, but it is
annabelx4

## Homework Statement

Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the earth.

In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy.

In this problem, you will apply the law of conservation of energy to different objects launched from the earth. The energy transformations that take place involve the object's kinetic energy and its gravitational potential energy . The law of conservation of energy for such cases implies that the sum of the object's kinetic energy and potential energy does not change with time. This idea can be expressed by the equation

K_i + U_1 + W_other = K_2 + U_2 ,

where "i" denotes the "initial" moment and "f" denotes the "final" moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem.

What is the speed of the object at the height of ?
Express your answer in terms of and . Use three significant figures in the numeric coefficient.

## Homework Equations

K_i + U_1 + W_other = K_2 + U_2

## The Attempt at a Solution

(1/2) mv^2 + 0 + 0 = (1/2) mv^2 + mg (v^2 / 4g)

so when I solve for v it = 0

mv^2 = 2(0)

v = 0

What did I do wrong?

So you're assuming it launches from the ground, so Ki=1/2*mVi^2

then at some other point it will have slowed down of course, so Kf=1/2*m*Vf^2, and the potential energy will be U=mgh

You tried to solve for h as a function of its velocity(which at that point will be the same Vf) as in the kinetic energy equation but I don't believe you did it right

You need to find that appropriate expression for the gravitational potential energy, U_1 and U_2, as functions of r, where r is the radius from the center of the mass responsible for the gravitational field.

http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html#gpt

## 1. What is the law of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, but can only be transformed from one form to another.

## 2. How does this law apply to an object?

This law applies to an object by stating that the total amount of energy within the object will remain constant, even as it undergoes changes in form or position.

## 3. What types of energy are involved in the conservation of energy?

There are many types of energy that can be involved in the conservation of energy, such as kinetic energy, potential energy, thermal energy, and chemical energy.

## 4. Can energy be transferred between objects?

Yes, energy can be transferred between objects. This is known as energy transfer and it occurs when energy is transferred from one object to another through physical contact or other means.

## 5. How does the law of conservation of energy impact everyday life?

The law of conservation of energy impacts everyday life in many ways, such as the energy we use to power our homes and vehicles, the food we eat for energy, and even the energy used in natural processes like photosynthesis and respiration.

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