A force of magnitude 2mg(1-ay) starts acting in the vertically upward direction on a body of mass m placed on earth's surface where y is the height of the object above the ground during ascent and 'a' is a positive constant. Total height through which the body ascends is

Options

A. `1//a`
B. `1//2a`
C. `2//3a`
D. `2//a`

Answer

To solve the problem, we need to determine the total height through which the body ascends when subjected to the force \( F = 2mg(1 - ay) \). Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Forces Acting on the Body
The body of mass \( m \) experiences two forces:
1. The gravitational force acting downwards, \( F_g = mg \).
2. The upward force given by \( F = 2mg(1 - ay) \).

### Step 2: Set Up the Equation for Equilibrium
At the point of equilibrium, the net force acting on the body will be zero. This occurs when the upward force equals the gravitational force:
\[ F = mg \]
Substituting the expression for \( F \):
\[ 2mg(1 - ay) = mg \]

### Step 3: Simplify the Equation
We can simplify the equation by dividing both sides by \( mg \) (assuming \( m \neq 0 \)):
\[ 2(1 - ay) = 1 \]
Now, solve for \( y \):
\[ 2 - 2ay = 1 \\ 2ay = 2 - 1 \\ 2ay = 1 \\ y = \frac{1}{2a} \]

### Step 4: Determine the Kinetic Energy at this Height
At the height \( y = \frac{1}{2a} \), the body has gained some kinetic energy. To find the maximum height, we use energy conservation principles. The kinetic energy at this point can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
where \( v \) is the velocity of the body at height \( y = \frac{1}{2a} \).

### Step 5: Use Energy Conservation to Find Maximum Height
At the maximum height \( h \), the kinetic energy will be zero, and all energy will be converted into potential energy. The potential energy at height \( h \) is given by:
\[ PE = mgh \]

Using conservation of mechanical energy:
\[ \text{Initial Energy} = \text{Final Energy} \]
At height \( \frac{1}{2a} \):
\[ \frac{1}{2} mv^2 + mg\left(\frac{1}{2a}\right) = mgh \]

### Step 6: Substitute the Velocity
From the previous steps, we can express \( v^2 \) in terms of the height:
\[ v^2 = 2g\left(\frac{1}{2a}\right) = \frac{g}{a} \]
Substituting \( v^2 \) into the energy conservation equation:
\[ \frac{1}{2} m \left(\frac{g}{a}\right) + mg\left(\frac{1}{2a}\right) = mgh \]

### Step 7: Simplify the Equation
\[ \frac{mg}{2a} + \frac{mg}{2a} = mgh \\ \frac{mg}{a} = mgh \]
Cancelling \( mg \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{a} = h \]

### Conclusion
Thus, the total height through which the body ascends is:
\[ h = \frac{1}{a} \]