Question

A particle moves under the action of a variable force \(F = (3x^2 - 2x + 5)\) in Newtons from \(x = 0\) to \(x = 2\) meters. The work done by the force is:

• A. \(10 \text{ J}\)
• B. \(12 \text{ J}\)
• C. \(14 \text{ J}\)
• D. \(16 \text{ J}\)

Hint:

For variable force calculations, look closely at the lower limit. If the lower limit is \(0\) and the expression is a simple polynomial, the entire lower limit evaluation becomes \(0\), saving you time during exams.

Answer 1

The Correct Option is C

Concept: When a force varies with position \(x\), the work done (\(W\)) cannot be computed using simple multiplication (\(F \cdot d\)). Instead, it must be determined by integrating the force function over the interval of displacement: \[ W = \int_{x_i}^{x_f} F \, dx \] We will apply standard polynomial integration rules: \(\int x^n \, dx = \frac{x^{n+1}}{n+1}\) and \(\int k \, dx = kx\). Step 1: Setting up the definite integral with the given limits.
The particle moves from the initial position \(x_i = 0\) to the final position \(x_f = 2\). Substituting the force function \(F = 3x^2 - 2x + 5\): \[ W = \int_{0}^{2} (3x^2 - 2x + 5) \, dx \]

Step 2: Integrating the expression term by term.
\[ W = \left[ 3\left(\frac{x^3}{3}\right) - 2\left(\frac{x^2}{2}\right) + 5x \right]_{0}^{2} \] Simplifying the coefficients: \[ W = \left[ x^3 - x^2 + 5x \right]_{0}^{2} \]

Step 3: Substituting the upper and lower limits.
Substitute the upper limit (\(x = 2\)): \[ W_{\text{upper}} = (2)^3 - (2)^2 + 5(2) = 8 - 4 + 10 = 14 \] Substitute the lower limit (\(x = 0\)): \[ W_{\text{lower}} = (0)^3 - (0)^2 + 5(0) = 0 \] Subtracting the lower limit value from the upper limit value: \[ W = 14 - 0 = 14 \text{ J} \]