When a spring a stretched through a distance x, it exerts a force given by `F=(-5x-16x^(3))N`. What is the work done, when the spring is stretched from 0.1 m to 0.2 m?

Options

A. `8.1xx10^(-1)J`
B. `12.2xx10^(-2)J`
C. `12.2xx10^(-1)J`
D. `8.1xx10^(-2)J`

Explanation

To find the work done when the spring is stretched from 0.1 m to 0.2 m, we can follow these steps:

### Step 1: Understand the Force Equation
The force exerted by the spring when stretched is given by:
\[ F = -5x - 16x^3 \]
This means that the force is a function of the displacement \( x \).

### Step 2: Set Up the Work Integral
The work done \( W \) on the spring when it is stretched from \( x_1 \) to \( x_2 \) can be calculated using the formula:
\[ W = -\int_{x_1}^{x_2} F \, dx \]
Since the force is acting in the opposite direction to the displacement, we have a negative sign.

### Step 3: Substitute the Force into the Integral
Substituting the expression for \( F \) into the integral, we get:
\[ W = -\int_{0.1}^{0.2} (-5x - 16x^3) \, dx \]
This simplifies to:
\[ W = \int_{0.1}^{0.2} (5x + 16x^3) \, dx \]

### Step 4: Integrate the Function
Now, we need to integrate the function:
\[ \int (5x + 16x^3) \, dx = \frac{5}{2}x^2 + 4x^4 + C \]
where \( C \) is the constant of integration.

### Step 5: Evaluate the Definite Integral
Now we will evaluate the definite integral from 0.1 to 0.2:
\[ W = \left[ \frac{5}{2}x^2 + 4x^4 \right]_{0.1}^{0.2} \]

Calculating the upper limit (0.2):
\[ W(0.2) = \frac{5}{2}(0.2)^2 + 4(0.2)^4 \]
\[ = \frac{5}{2}(0.04) + 4(0.0016) \]
\[ = 0.1 + 0.0064 = 0.1064 \]

Calculating the lower limit (0.1):
\[ W(0.1) = \frac{5}{2}(0.1)^2 + 4(0.1)^4 \]
\[ = \frac{5}{2}(0.01) + 4(0.0001) \]
\[ = 0.025 + 0.0004 = 0.0254 \]

### Step 6: Subtract the Lower Limit from the Upper Limit
Now, we subtract the lower limit from the upper limit:
\[ W = W(0.2) - W(0.1) \]
\[ = 0.1064 - 0.0254 = 0.081 \, \text{J} \]

### Final Answer
The work done when the spring is stretched from 0.1 m to 0.2 m is:
\[ W = 0.081 \, \text{J} \]