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. `8.7xx10^(-2)J`
B. `12.2xx10^(-2)J`
C. `8.7xx10^(-1)J`
D. `12.2xx10^(-1)J`

Answer

To find the work done when the spring is stretched from 0.1 m to 0.2 m, we start with the given force equation for the spring:

\[ F = -5x - 16x^3 \]

### Step 1: Understand the Work Done
The work done \( W \) on the spring when stretching it from position \( x_1 \) to \( x_2 \) is given by the integral of the force over the distance:

\[ W = \int_{x_1}^{x_2} F \, dx \]

### Step 2: Set the Limits
In our case, we are stretching the spring from \( x_1 = 0.1 \, \text{m} \) to \( x_2 = 0.2 \, \text{m} \). Therefore, we can write:

\[ W = \int_{0.1}^{0.2} (-5x - 16x^3) \, dx \]

### Step 3: Calculate the Integral
Now we need to compute the integral:

\[ W = \int_{0.1}^{0.2} (-5x - 16x^3) \, dx \]

We can split the integral:

\[ W = \int_{0.1}^{0.2} -5x \, dx + \int_{0.1}^{0.2} -16x^3 \, dx \]

Calculating each integral separately:

1. **First Integral:**
\[ \int -5x \, dx = -\frac{5}{2} x^2 \]
Evaluating from 0.1 to 0.2:
\[ = -\frac{5}{2} (0.2^2) + \frac{5}{2} (0.1^2) = -\frac{5}{2} (0.04) + \frac{5}{2} (0.01) = -0.1 + 0.025 = -0.075 \]

2. **Second Integral:**
\[ \int -16x^3 \, dx = -4x^4 \]
Evaluating from 0.1 to 0.2:
\[ = -4(0.2^4) + 4(0.1^4) = -4(0.0016) + 4(0.0001) = -0.0064 + 0.0004 = -0.006 \]

### Step 4: Combine the Results
Now we combine the results of both integrals:

\[ W = -0.075 - 0.006 = -0.081 \, \text{J} \]

### Step 5: Final Calculation
Thus, the total work done in stretching the spring from 0.1 m to 0.2 m is:

\[ W = -0.081 \, \text{J} \]

### Conclusion
The work done when the spring is stretched from 0.1 m to 0.2 m is approximately:

\[ W \approx -0.081 \, \text{J} \]