A block of mass `5 kg` initially at rest at the origin is acted upon by a force along the positive `X-direction` represented by
`F=(20+5x) N`. Calculate the work done by the force during the displacement of the block from `x = 0` to `x = 4m`.

Options

A. 100 J
B. 150 J
C. 120 J
D. 75 J

Explanation

To calculate the work done by the force on the block as it moves from \( x = 0 \) to \( x = 4 \, m \), we can follow these steps:

### Step 1: Identify the Force Function
The force acting on the block is given by:
\[ F(x) = 20 + 5x \quad \text{(in Newtons)} \]

### Step 2: Set Up the Work Done Integral
The work done \( W \) by a variable force as the object moves from position \( x_1 \) to \( x_2 \) is given by:
\[ W = \int_{x_1}^{x_2} F(x) \, dx \]
In this case, \( x_1 = 0 \) and \( x_2 = 4 \).

### Step 3: Substitute the Force Function into the Integral
Substituting the force function into the integral, we have:
\[ W = \int_{0}^{4} (20 + 5x) \, dx \]

### Step 4: Calculate the Integral
We can split the integral into two parts:
\[ W = \int_{0}^{4} 20 \, dx + \int_{0}^{4} 5x \, dx \]

Calculating the first integral:
\[ \int_{0}^{4} 20 \, dx = 20x \bigg|_{0}^{4} = 20 \times 4 - 20 \times 0 = 80 \]

Calculating the second integral:
\[ \int_{0}^{4} 5x \, dx = \frac{5}{2} x^2 \bigg|_{0}^{4} = \frac{5}{2} (4^2) - \frac{5}{2} (0^2) = \frac{5}{2} \times 16 = 40 \]

### Step 5: Add the Results of the Integrals
Now, we add the results of the two integrals:
\[ W = 80 + 40 = 120 \, \text{Joules} \]

### Final Answer
The work done by the force during the displacement of the block from \( x = 0 \) to \( x = 4 \, m \) is:
\[ \boxed{120 \, \text{Joules}} \]