How much work is required to move an object with a force of $F\left(x\right)=4x^{3}N$ acting along the $x$-axis from $x=0m$ to $x=2m$?

• A. $48J$
• B. $$$16J$
• C. $64J$
• D. $4J$

Answer: (B)

Verified step by step guidance

1. Step 1: Recall the formula for work done by a variable force along a straight line: \( W = \int_{a}^{b} F(x) \, dx \), where \( F(x) \) is the force function, and \( a \) and \( b \) are the limits of integration.
2. Step 2: Identify the given force function \( F(x) = 4x^3 \) and the limits of integration \( x = 0 \) to \( x = 2 \). Substitute these values into the formula for work.
3. Step 3: Set up the integral: \( W = \int_{0}^{2} 4x^3 \, dx \). Factor out the constant \( 4 \) to simplify the integral: \( W = 4 \int_{0}^{2} x^3 \, dx \).
4. Step 4: Use the power rule for integration: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. Apply this rule to \( x^3 \): \( \int x^3 \, dx = \frac{x^4}{4} \).
5. Step 5: Evaluate the definite integral by substituting the limits \( x = 0 \) and \( x = 2 \) into \( \frac{x^4}{4} \). Compute \( W = 4 \left[ \frac{2^4}{4} - \frac{0^4}{4} \right] \). Simplify the expression to find the work done.