A force F is related to the position of a particle by the relation `F=(10x^(2))N` . Find the work done by the force when the particle moves from `x=2m to x=4m`.

Options

A. `56/3` J
B. 560 J
C. `560/3` J
D. `3/560` J

Answer

To find the work done by the force when the particle moves from \( x = 2 \, \text{m} \) to \( x = 4 \, \text{m} \), we can follow these steps:

### Step 1: Understand the relationship between force and position
The force \( F \) is given by the relation:
\[ F = 10x^2 \, \text{N} \]
where \( x \) is the position in meters.

### Step 2: Set up the work done integral
The work done \( W \) by a variable force is calculated using the integral:
\[ W = \int_{x_1}^{x_2} F \, dx \]
In this case, \( x_1 = 2 \, \text{m} \) and \( x_2 = 4 \, \text{m} \). Thus, we can write:
\[ W = \int_{2}^{4} 10x^2 \, dx \]

### Step 3: Calculate the integral
Now, we need to evaluate the integral:
\[ W = 10 \int_{2}^{4} x^2 \, dx \]
To solve this integral, we use the power rule for integration:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]
For \( n = 2 \):
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Now substituting this back into our work equation:
\[ W = 10 \left[ \frac{x^3}{3} \right]_{2}^{4} \]

### Step 4: Evaluate the definite integral
Now we evaluate the definite integral from \( 2 \) to \( 4 \):
\[ W = 10 \left( \frac{4^3}{3} - \frac{2^3}{3} \right) \]
Calculating \( 4^3 \) and \( 2^3 \):
\[ 4^3 = 64 \quad \text{and} \quad 2^3 = 8 \]
Substituting these values:
\[ W = 10 \left( \frac{64}{3} - \frac{8}{3} \right) = 10 \left( \frac{64 - 8}{3} \right) = 10 \left( \frac{56}{3} \right) \]

### Step 5: Final calculation
Now multiply:
\[ W = \frac{560}{3} \, \text{J} \]

### Conclusion
Thus, the work done by the force when the particle moves from \( x = 2 \, \text{m} \) to \( x = 4 \, \text{m} \) is:
\[ W = \frac{560}{3} \, \text{J} \]