A force `f =10 + 2 x` acts on a particle moving in straight line on x axis the work done by this force during a displacement from `x =0` and `x =3m`

Options

A. `13`J
B. `29`J
C. `39`J
D. `43`J

Answer

To find the work done by the force \( f = 10 + 2x \) during the displacement from \( x = 0 \) to \( x = 3 \) m, we will follow these steps:

### Step 1: Understand the Work Done Formula
The work done \( W \) by a variable force is given by the integral of the force over the displacement:
\[ W = \int_{x_1}^{x_2} f(x) \, dx \]
where \( f(x) \) is the force as a function of position \( x \), and \( x_1 \) and \( x_2 \) are the initial and final positions.

### Step 2: Set Up the Integral
Given the force \( f = 10 + 2x \), we need to set up the integral from \( x = 0 \) to \( x = 3 \):
\[ W = \int_{0}^{3} (10 + 2x) \, dx \]

### Step 3: Calculate the Integral
Now we will compute the integral:
\[ W = \int_{0}^{3} (10 + 2x) \, dx = \int_{0}^{3} 10 \, dx + \int_{0}^{3} 2x \, dx \]

Calculating each part:
1. For \( \int_{0}^{3} 10 \, dx \):
\[ \int 10 \, dx = 10x \Big|_{0}^{3} = 10(3) - 10(0) = 30 \]

2. For \( \int_{0}^{3} 2x \, dx \):
\[ \int 2x \, dx = x^2 \Big|_{0}^{3} = (3^2) - (0^2) = 9 \]

Adding both results together:
\[ W = 30 + 9 = 39 \, \text{Joules} \]

### Step 4: Conclusion
The work done by the force during the displacement from \( x = 0 \) m to \( x = 3 \) m is:
\[ W = 39 \, \text{J} \]