A force F=a+bx acts on a particle in the x-directioin, where a and b are constants. Find the work done by this force during a displacement form `x=0 to x=d`.

Explanation

To find the work done by the force \( F = a + bx \) during the displacement from \( x = 0 \) to \( x = d \), we will follow these steps:

### Step 1: Understand the Work Done by a Variable Force
The work done \( W \) by a force \( F \) when it acts over a displacement \( dx \) is given by the integral:
\[ W = \int F \, dx \]
Since the force \( F \) is a function of \( x \), we will need to integrate it over the specified limits.

### Step 2: Set Up the Integral
Given the force \( F = a + bx \), we can express the work done as:
\[ W = \int_{0}^{d} (a + bx) \, dx \]

### Step 3: Perform the Integration
Now, we will integrate the expression:
\[ W = \int_{0}^{d} (a + bx) \, dx = \int_{0}^{d} a \, dx + \int_{0}^{d} bx \, dx \]

Calculating each integral separately:
1. For the first integral:
\[ \int_{0}^{d} a \, dx = a \cdot x \bigg|_{0}^{d} = a \cdot d - a \cdot 0 = ad \]

2. For the second integral:
\[ \int_{0}^{d} bx \, dx = b \cdot \frac{x^2}{2} \bigg|_{0}^{d} = b \cdot \frac{d^2}{2} - b \cdot \frac{0^2}{2} = \frac{bd^2}{2} \]

### Step 4: Combine the Results
Now, we can combine the results of both integrals:
\[ W = ad + \frac{bd^2}{2} \]

### Step 5: Final Answer
Thus, the work done by the force \( F \) during the displacement from \( x = 0 \) to \( x = d \) is:
\[ W = ad + \frac{bd^2}{2} \]