A particle moves along x - axis under the action of a position dependent force `F=(5x^(2)-2x)N`. Work done by force on the particle when it moves from origin to x = 3 m is

Options

A. 45 J
B. 36 J
C. 32 J
D. 42 J

Answer

To solve the problem of calculating the work done by the force \( F = (5x^2 - 2x) \, \text{N} \) on a particle moving from the origin (0 m) to \( x = 3 \, \text{m} \), we will follow these steps:

### Step 1: Understand the Work Done by a Variable Force
The work done \( W \) by a variable force when moving an object from position \( x_1 \) to \( x_2 \) is given by the integral of the force over the displacement:
\[ W = \int_{x_1}^{x_2} F(x) \, dx \]

### Step 2: Set Up the Integral
In this case, we need to calculate the work done from \( x = 0 \) to \( x = 3 \):
\[ W = \int_{0}^{3} (5x^2 - 2x) \, dx \]

### Step 3: Calculate the Integral
Now, we will compute the integral:
1. Integrate \( 5x^2 \):
\[ \int 5x^2 \, dx = \frac{5}{3} x^3 \]
2. Integrate \( -2x \):
\[ \int -2x \, dx = -x^2 \]
3. Combine the results:
\[ W = \left[ \frac{5}{3} x^3 - x^2 \right]_{0}^{3} \]

### Step 4: Evaluate the Integral at the Limits
Now we will evaluate the expression at the limits \( x = 3 \) and \( x = 0 \):
1. At \( x = 3 \):
\[ W = \frac{5}{3} (3^3) - (3^2) = \frac{5}{3} (27) - 9 = \frac{135}{3} - 9 = 45 - 9 = 36 \]
2. At \( x = 0 \):
\[ W = \frac{5}{3} (0^3) - (0^2) = 0 - 0 = 0 \]

### Step 5: Calculate the Work Done
Thus, the total work done when the particle moves from \( x = 0 \) to \( x = 3 \) is:
\[ W = 36 \, \text{Joules} \]

### Final Answer
The work done by the force on the particle when it moves from the origin to \( x = 3 \, \text{m} \) is \( 36 \, \text{J} \).

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