A particle starts with speed `v_(0)` from x = 0 along x - axis with retardation proportional to the square of its displacement. Work done by the force acting on the particle is proportional to

Options

A. `x^((5)/(2))`
B. `x^(3)`
C. `e^(x)`
D. `x^(2)`

Answer

To solve the problem, we need to analyze the motion of a particle that experiences retardation proportional to the square of its displacement. We'll follow these steps:

### Step 1: Define the retardation
The problem states that the retardation (deceleration) is proportional to the square of the displacement. We can express this mathematically as:
\[ a = -k x^2 \]
where \( k \) is a proportionality constant, \( a \) is the acceleration (which is negative due to retardation), and \( x \) is the displacement.

### Step 2: Relate acceleration to force
Since acceleration is related to force by Newton's second law, we can express the force \( F \) acting on the particle as:
\[ F = m a = -m k x^2 \]
where \( m \) is the mass of the particle.

### Step 3: Work done by the force
The work done \( dW \) by the force when the particle moves through a small displacement \( dx \) is given by:
\[ dW = F \, dx = -m k x^2 \, dx \]

### Step 4: Integrate to find total work done
To find the total work done \( W \) as the particle moves from \( x = 0 \) to \( x \), we need to integrate:
\[ W = \int_0^x dW = \int_0^x (-m k x^2) \, dx \]
This simplifies to:
\[ W = -m k \int_0^x x^2 \, dx \]

### Step 5: Calculate the integral
The integral of \( x^2 \) is:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Thus, we have:
\[ W = -m k \left[ \frac{x^3}{3} \right]_0^x = -m k \frac{x^3}{3} \]

### Step 6: Determine the proportionality
From the expression for work done, we can see that:
\[ W \propto -x^3 \]
Since we are interested in the proportionality, we can ignore the negative sign and the constants, leading us to conclude:
\[ W \propto x^3 \]

### Conclusion
The work done by the force acting on the particle is proportional to \( x^3 \).