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 Formula
The work done \( W \) by a force \( F \) when moving an object through a displacement \( ds \) is given by the integral:
\[ W = \int F \cdot ds \]
Since the force is acting in the y-direction and the displacement is also in the y-direction, we can write:
\[ W = \int F \, dy \]

### Step 2: Substitute the Force Function
We know that the force \( F \) is given by:
\[ F = 20 + 10y \]
Thus, we can substitute this into the work formula:
\[ W = \int (20 + 10y) \, dy \]

### Step 3: Set the Limits of Integration
The particle moves from \( y = 0 \) to \( y = 1 \) meter. Therefore, we set the limits of integration:
\[ W = \int_{0}^{1} (20 + 10y) \, dy \]

### Step 4: Perform the Integration
Now we can integrate the function:
\[ W = \int_{0}^{1} (20 + 10y) \, dy = \left[ 20y + 5y^2 \right]_{0}^{1} \]
Here, we integrated \( 20y \) to get \( 20y \) and \( 10y \) to get \( 5y^2 \) (since \( \int 10y \, dy = 5y^2 \)).

### Step 5: Evaluate the Integral at the Limits
Now we will evaluate the integral from \( 0 \) to \( 1 \):
\[ W = \left[ 20(1) + 5(1)^2 \right] - \left[ 20(0) + 5(0)^2 \right] \]
Calculating this gives:
\[ W = (20 + 5) - (0 + 0) = 25 - 0 = 25 \text{ Joules} \]

### Conclusion
The work done by the force when moving the particle from \( y = 0 \) to \( y = 1 \) meter is \( 25 \) Joules.

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