A force `F = - kx^(3)` is acting on a block moving along x-axis. Here, k is a positive constant. Work done by this force is

Options

A. positive in displacing the block from x = 3 to x = 1
B. positive in displacing the block from x = - 1 to x = - 3
C. negative in displacing the block from x = 3 to x = 1
D. negative in displacing the block from x = - 1 to x = - 3

Answer

To find the work done by the force \( F = -kx^3 \) acting on a block moving along the x-axis, we can follow these steps:

### Step 1: Understand the Work Done
The work done \( W \) by a force when an object moves from position \( x_i \) to position \( x_f \) is given by the integral of the force over the displacement:
\[ W = \int_{x_i}^{x_f} F \, dx \]

### Step 2: Substitute the Force
We substitute the expression for the force \( F = -kx^3 \) into the work done formula:
\[ W = \int_{x_i}^{x_f} -kx^3 \, dx \]

### Step 3: Integrate the Force
Now we perform the integration:
\[ W = -k \int_{x_i}^{x_f} x^3 \, dx \]
The integral of \( x^3 \) is:
\[ \int x^3 \, dx = \frac{x^4}{4} \]
Thus, we can write:
\[ W = -k \left[ \frac{x^4}{4} \right]_{x_i}^{x_f} \]

### Step 4: Evaluate the Integral
Now we evaluate the definite integral:
\[ W = -k \left( \frac{x_f^4}{4} - \frac{x_i^4}{4} \right) \]
This simplifies to:
\[ W = -\frac{k}{4} \left( x_f^4 - x_i^4 \right) \]

### Step 5: Final Expression for Work Done
Thus, the work done by the force \( F = -kx^3 \) when moving from \( x_i \) to \( x_f \) is:
\[ W = -\frac{k}{4} \left( x_f^4 - x_i^4 \right) \]