A metal can is pushed from $x=0.5m$ to $x=1.5m$ with a force given by the function $F\left(x\right)=2e^{-2x}$. How much work is done by the force on the metal can?

• A. $0.636J$
• B. $0.318J$
• C. $1.214J$
• D. $0.050J$

Answer: (B)

Verified step by step guidance

1. Identify the physical quantity to find: The problem asks for the work done by a variable force on the metal can as it moves from an initial position to a final position.
2. Recall the formula for work done by a variable force along a straight line: \(W = \int_{x_i}^{x_f} F(x) \, dx\), where \(x_i\) and \(x_f\) are the initial and final positions, respectively.
3. Substitute the given force function and limits into the integral: \(W = \int_0^{0.5} 2 e^{-2x} \, dx\).
4. Set up the integral for evaluation: Recognize that the integral of \(e^{ax}\) is \(\frac{1}{a} e^{ax}\), so you will integrate \(2 e^{-2x}\) accordingly.
5. Evaluate the definite integral by calculating the antiderivative at the upper and lower limits and subtracting: \(W = \left[ -e^{-2x} \right]_0^{0.5}\).