A force `F=20+10y` acts on a particle in y-direction where F is in Newton and y in meter. Wrok done by this force to move the particle from `y=0` to `y=1 m` is:

Options

A. 20 J
B. 30J
C. 5J
D. 25J

Explanation

To find the work done by the force \( F = 20 + 10y \) when moving a particle from \( y = 0 \) to \( y = 1 \) meter, we can follow these steps:

### Step 1: Understand the Work Done by a Variable Force
The work done \( W \) by a variable force is calculated using the integral of the force over the distance moved. The formula is given by:
\[ W = \int_{y_1}^{y_2} F \, dy \]
where \( y_1 \) and \( y_2 \) are the initial and final positions, respectively.

### Step 2: Set Up the Integral
In this case, the force \( F \) is given as \( 20 + 10y \). We need to calculate the work done from \( y = 0 \) to \( y = 1 \):
\[ W = \int_{0}^{1} (20 + 10y) \, dy \]

### Step 3: Perform the Integration
Now, we can integrate the force function:
\[ W = \int_{0}^{1} (20 + 10y) \, dy = \int_{0}^{1} 20 \, dy + \int_{0}^{1} 10y \, dy \]

Calculating each integral separately:
1. For \( \int_{0}^{1} 20 \, dy \):
\[ \int 20 \, dy = 20y \bigg|_{0}^{1} = 20(1) - 20(0) = 20 \]

2. For \( \int_{0}^{1} 10y \, dy \):
\[ \int 10y \, dy = 10 \cdot \frac{y^2}{2} \bigg|_{0}^{1} = 5y^2 \bigg|_{0}^{1} = 5(1^2) - 5(0^2) = 5 \]

### Step 4: Combine the Results
Now, we can combine the results of the two integrals:
\[ W = 20 + 5 = 25 \, \text{Joules} \]

### Final Answer
Thus, the work done by the force as the particle moves from \( y = 0 \) to \( y = 1 \) meter is:
\[ \boxed{25 \, \text{J}} \]