Question

A particle is subjected to a force F = 3x$^2$. Calculate the work done in moving with particle from x=0 to x = 2 m.

• A. 12 J
• B. 8 J
• C. 16 J

Hint:

Remember the basic integration rule: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$. For variable forces, always integrate $F(x)dx$ over the specified limits.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Work is a measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement. When the force is not constant but varies with position (as given here, \(F = 3x^2\)), we must use integration to calculate the work done.
Step 2: Key Formula or Approach:
The general formula for work done by a variable force \(F(x)\) from an initial position \(x_i\) to a final position \(x_f\) is:
\[ W = \int_{x_i}^{x_f} F(x) \,dx \]
Step 3: Detailed Explanation:
Given force function: \(F(x) = 3x^2\) N
Initial position: \(x_i = 0\) m
Final position: \(x_f = 2\) m
Set up the definite integral for work:
\[ W = \int_{0}^{2} 3x^2 \,dx \]
To evaluate this integral, find the antiderivative of \(3x^2\). The antiderivative of \(x^n\) is \(\frac{x^{n+1}}{n+1}\).
So, the antiderivative of \(3x^2\) is \(3 \cdot \frac{x^3}{3} = x^3\).
Now apply the limits of integration from 0 to 2:
\[ W = [x^3]_{0}^{2} \]
\[ W = (2)^3 - (0)^3 \]
\[ W = 8 - 0 \]
\[ W = 8 \text{ Joules} \]
Step 4: Final Answer:
The work done in moving the particle is 8 J.