Question

A particle moves along the x-axis under a force $ F(x) = 6x^2 $ N. The work done by this force in moving the particle from $ x = 1 \, \text{m} $ to $ x = 2 \, \text{m} $ is:

• A. 14 J
• B. 18 J
• C. 24 J
• D. 28 J

Hint:

When force is a function of position, use definite integration: \[ W = \int_{x_1}^{x_2} F(x)\,dx \]

Answer 1

The Correct Option is A

To find the work done by the force $ F(x) = 6x^2 $ as the particle moves from $ x = 1 \, \text{m} $ to $ x = 2 \, \text{m} $, we use the work integral formula:
\[ W = \int_{x_1}^{x_2} F(x) \, dx \] Substituting the given values, we have: \[ W = \int_{1}^{2} 6x^2 \, dx \] Evaluate the integral: \[ W = 6 \int_{1}^{2} x^2 \, dx \] We know that the integral of $ x^n $ is $ \frac{x^{n+1}}{n+1} $. Applying this, we get: \[ \int x^2 \, dx = \frac{x^3}{3} \] Therefore: \[ W = 6 \left[ \frac{x^3}{3} \right]_{1}^{2} \] Simplifying: \[ W = 6 \left( \frac{2^3}{3} - \frac{1^3}{3} \right) \] Calculate: \[ W = 6 \left( \frac{8}{3} - \frac{1}{3} \right) = 6 \left( \frac{7}{3} \right) = 2 \times 7 = 14 \, \text{J} \] Thus, the work done by the force is 14 J.

Answer 2

Work done by a variable force along the x-direction is given by: \[ W = \int_{x_1}^{x_2} F(x) \, dx \] Given \( F(x) = 6x^2 \), and limits: \( x_1 = 1 \), \( x_2 = 2 \) Evaluate the definite integral: \[ W = \int_{1}^{2} 6x^2 \, dx \] Step 1: Factor out the constant \[ W = 6 \int_{1}^{2} x^2 \, dx \] Step 2: Integrate the function \[ \int x^2 \, dx = \frac{x^3}{3} \] \[ W = 6 \left[ \frac{x^3}{3} \right]_1^2 \] Step 3: Apply the limits \[ \left[ \frac{x^3}{3} \right]_1^2 = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} \] Step 4: Multiply by the constant \[ W = 6 \cdot \frac{7}{3} = \frac{42}{3} = 14 \] Final Answer: \[ \boxed{W = 14 \text{ J}} \]