A force `F = Ay^(2)+By+C` acts on a body at rest in the `Y`-direction. The kinetic energy of the body during a displacement `y = -a` to `y = a` is

Options

A. `(2Aa^(3))/(3)`
B. `(2Aa^(3))/(3)+2Ca`
C. `(2Aa^(3))/(3)+(Ba^(2))/(2)+Ca`
D. None of these

Explanation

To find the kinetic energy of a body subjected to a force given by \( F = Ay^2 + By + C \) during a displacement from \( y = -a \) to \( y = a \), we can follow these steps:

### Step 1: Understand the relationship between work and kinetic energy
According to the work-energy theorem, the work done by all forces on an object is equal to the change in kinetic energy of that object. Mathematically, this is expressed as:
\[ W = \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \]
Since the body starts at rest, the initial kinetic energy \( KE_{\text{initial}} = 0 \). Therefore, the work done will be equal to the final kinetic energy:
\[ W = KE_{\text{final}} \]

### Step 2: Write the expression for work done
The work done by a force \( F \) when moving an object through a displacement \( dy \) is given by the integral:
\[ W = \int F \, dy \]
Substituting the expression for the force:
\[ W = \int (Ay^2 + By + C) \, dy \]

### Step 3: Set the limits of integration
We need to evaluate this integral from \( y = -a \) to \( y = a \):
\[ W = \int_{-a}^{a} (Ay^2 + By + C) \, dy \]

### Step 4: Evaluate the integral
Now we calculate the integral:
\[ W = \int_{-a}^{a} (Ay^2 + By + C) \, dy = \left[ \frac{A y^3}{3} + \frac{B y^2}{2} + Cy \right]_{-a}^{a} \]

### Step 5: Substitute the limits into the integral
Now we substitute the limits into the evaluated integral:
\[ W = \left( \frac{A a^3}{3} + \frac{B a^2}{2} + Ca \right) - \left( \frac{A (-a)^3}{3} + \frac{B (-a)^2}{2} + C(-a) \right) \]
This simplifies to:
\[ W = \left( \frac{A a^3}{3} + \frac{B a^2}{2} + Ca \right) - \left( -\frac{A a^3}{3} + \frac{B a^2}{2} - Ca \right) \]

### Step 6: Simplify the expression
Combining the terms:
\[ W = \frac{A a^3}{3} + \frac{B a^2}{2} + Ca + \frac{A a^3}{3} - \frac{B a^2}{2} + Ca \]
\[ W = \frac{2A a^3}{3} + 2Ca \]

### Step 7: Relate work done to kinetic energy
Since the work done is equal to the change in kinetic energy, we have:
\[ KE_{\text{final}} = W = \frac{2A a^3}{3} + 2Ca \]

### Final Answer
Thus, the kinetic energy of the body during the displacement from \( y = -a \) to \( y = a \) is:
\[ KE_{\text{final}} = \frac{2A a^3}{3} + 2Ca \]