Problem Statement
Solve the work-energy problem: A force $F = (3x^2 + 2x)$ N acts on a particle. Find the work done as the particle moves from $x = 0$ to $x = 2$ m. $W = \int_{x_1}^{x_2} F\,dx$ Step 1: $W = \displaystyle\int_0^2 (3x^2 + 2x)\,dx = \left[x^3 + x^2\right]_0^2 = (8+4) – 0 = 12$ J. $$\boxed{W = 12\text{ J}}$$
Given Information
- $\boxed{W = 12\text{ J}$
Physical Concepts & Formulas
Newton’s second law $\mathbf{F}_\text{net} = m\mathbf{a}$ is the fundamental relation between net force and acceleration. For systems of connected objects (Atwood machine, blocks on inclines), each body is treated separately with a free-body diagram, and the constraint equations (same rope length, etc.) link the accelerations.
- $\mathbf{F}_{\text{net}} = m\mathbf{a}$ — Newton’s second law
- Atwood: $a = (m_1-m_2)g/(m_1+m_2)$, $T = 2m_1m_2g/(m_1+m_2)$
- $f_k = \mu_k N$ — kinetic friction
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\boxed{W = 12\text{ J}}$$
$$\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f$$
$$E_i – W_{\text{friction}} = E_f \implies \frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f + f_k d$$
Answer
$$\boxed{W = 12\text{ J}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
Leave a Reply