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 determine the work done by the force acting on a particle that experiences retardation proportional to the square of its displacement. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Problem
The particle starts with an initial speed \( v_0 \) from \( x = 0 \) and experiences retardation that is proportional to the square of its displacement \( x \). This means we can express the retardation (deceleration) as:
\[ a = -k x^2 \]
where \( k \) is a proportionality constant.

### Step 2: Relate Force and Acceleration
Using Newton's second law, we know that:
\[ F = ma \]
Since the mass \( m \) is constant, we can express the force as:
\[ F = m(-k x^2) = -mk x^2 \]
Thus, the force acting on the particle is proportional to \( -x^2 \).

### Step 3: Work Done by the Force
The work done \( W \) by the force when the particle moves from an initial position to a final position can be expressed as:
\[ W = \int F \, dx \]
Substituting the expression for force:
\[ W = \int_{0}^{x} -mk x^2 \, dx \]

### Step 4: Perform the Integration
Now we can perform the integration:
\[ W = -mk \int_{0}^{x} x^2 \, dx \]
The integral of \( x^2 \) is:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Thus, we have:
\[ W = -mk \left[ \frac{x^3}{3} \right]_{0}^{x} = -mk \left( \frac{x^3}{3} - 0 \right) = -\frac{mk x^3}{3} \]

### Step 5: Determine Proportionality
Since we are interested in the proportionality of the work done \( W \), we can express it as:
\[ W \propto -x^3 \]
Ignoring the negative sign (as we are only interested in the proportionality), we conclude that:
\[ W \propto x^3 \]

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

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