A force `F=Kx^(2)` acts on a particle at an angle of `60°` with the x–axis. the work done in displacing the particle from `x_(1)` to `x_(2)` will be –

Options

A. `(kx^(2))/(2)`
B. `(k)/(2)(x_(2)^(2)-x_(1)^(2))`
C. `(k)/(6)(x_(2)^(3)-x_(1)^(3))`
D. `(k)/(3)(x_(2)^(3)-x_(1)^(3))`

Answer

To solve the problem of calculating the work done by the force \( F = Kx^2 \) acting at an angle of \( 60^\circ \) with the x-axis while displacing the particle from \( x_1 \) to \( x_2 \), we can follow these steps:

### Step 1: Identify the Components of the Force
The force \( F \) acts at an angle of \( 60^\circ \) to the x-axis. The x-component of the force is given by:
\[ F_x = F \cos(60^\circ) = Kx^2 \cdot \frac{1}{2} = \frac{Kx^2}{2} \]

### Step 2: Write the Expression for Work Done
The work done \( dW \) by the force when the particle is displaced by an infinitesimal distance \( dx \) is given by:
\[ dW = F_x \cdot dx = \frac{Kx^2}{2} \cdot dx \]

### Step 3: Integrate to Find Total Work Done
To find the total work done in displacing the particle from \( x_1 \) to \( x_2 \), we need to integrate \( dW \):
\[ W = \int_{x_1}^{x_2} dW = \int_{x_1}^{x_2} \frac{Kx^2}{2} \, dx \]

### Step 4: Perform the Integration
Now we can perform the integration:
\[ W = \frac{K}{2} \int_{x_1}^{x_2} x^2 \, dx \]
The integral of \( x^2 \) is:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Thus, we have:
\[ W = \frac{K}{2} \left[ \frac{x^3}{3} \right]_{x_1}^{x_2} \]

### Step 5: Substitute the Limits
Now substituting the limits \( x_1 \) and \( x_2 \):
\[ W = \frac{K}{2} \left( \frac{x_2^3}{3} - \frac{x_1^3}{3} \right) \]
This simplifies to:
\[ W = \frac{K}{6} (x_2^3 - x_1^3) \]

### Final Answer
Thus, the work done in displacing the particle from \( x_1 \) to \( x_2 \) is:
\[ W = \frac{K}{6} (x_2^3 - x_1^3) \]

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