When a body falls freely towards the earth, then its total energy

Options

A. increases
B. decreases
C. remains constant
D. first increases and then decreases

Explanation

To solve the question regarding the total energy of a body falling freely towards the Earth, we can break it down into several steps:

### Step 1: Understand the Concept of Total Energy
The total mechanical energy of a body is the sum of its kinetic energy (KE) and potential energy (PE). The formula for total energy (E) can be expressed as:
\[ E = KE + PE \]

### Step 2: Identify the Initial Conditions
When a body is at a height \( h \) and is released (not thrown), its initial velocity \( v \) is 0. Therefore, the kinetic energy at this point is:
\[ KE = \frac{1}{2} m v^2 = \frac{1}{2} m (0)^2 = 0 \]

The potential energy at this height is given by:
\[ PE = mgh \]
Thus, the total energy at the initial point (let's call it point A) is:
\[ E_A = KE + PE = 0 + mgh = mgh \]

### Step 3: Analyze the Energy at a Lower Height
As the body falls freely, it loses potential energy and gains kinetic energy. Let's consider the body when it has fallen to a height of \( \frac{h}{2} \) (let's call this point B).

At point B:
- The height is \( \frac{h}{2} \), so the potential energy is:
\[ PE_B = mg\left(\frac{h}{2}\right) = \frac{mgh}{2} \]

- To find the kinetic energy at this point, we can use the conservation of energy principle. Since total energy must remain constant, we have:
\[ E_B = E_A = mgh \]
Thus, we can express the kinetic energy at point B as:
\[ KE_B = E_B - PE_B = mgh - \frac{mgh}{2} = \frac{mgh}{2} \]

### Step 4: Calculate the Total Energy at Point B
Now, we can find the total energy at point B:
\[ E_B = KE_B + PE_B = \frac{mgh}{2} + \frac{mgh}{2} = mgh \]

### Step 5: Conclude the Total Energy
From the analysis at points A and B, we see that the total energy remains constant throughout the fall:
\[ E = mgh \text{ (constant)} \]

### Final Statement
Therefore, the total energy of a body falling freely towards the Earth remains constant and is equal to \( mgh \) at any point during the fall.

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