Problem Statement
Solve the work-energy problem: Solve the work-energy problem: Spring of stiffness k stretched by x₀, mass m released. Find KE when spring reaches natural length. Mass m, spring constant k, initial extension x₀ $$KE + PE = E_{total} = \tfrac{1}{2}kx_0^2$$ Step 1: Initial energy: E = ½kx₀² (all PE). Step 2: At natural length: PE =
Given Information
- Mass $m$, velocity $v$, height $h$, or other given quantities
- Any forces doing work (conservative or non-conservative) as specified
Physical Concepts & Formulas
The Work-Energy Theorem states that the net work done on an object equals its change in kinetic energy: $W_{\text{net}} = \Delta KE$. For conservative forces (gravity, spring, electric), a potential energy function $U$ exists such that $W = -\Delta U$, and the total mechanical energy $E = KE + U$ is conserved. Non-conservative forces (friction, air drag) remove mechanical energy, converting it to thermal energy. The power delivered is $P = dW/dt = \vec{F}\cdot\vec{v}$.
- $W = \vec{F}\cdot\vec{d} = Fd\cos\theta$ — work done by constant force
- $KE = \frac{1}{2}mv^2$ — kinetic energy
- $U_g = mgh$ — gravitational PE (near Earth’s surface)
- $U_s = \frac{1}{2}kx^2$ — elastic PE
- $W_{\text{net}} = \Delta KE = KE_f – KE_i$ — work-energy theorem
- $E_i = E_f$ (when only conservative forces act)
Step-by-Step Solution
Step 1 — Identify all forces and whether they are conservative.
Step 2 — Apply conservation of energy (if no friction):
$$\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f$$
Step 3 — If friction acts:
$$E_i – W_{\text{friction}} = E_f \implies \frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f + f_k d$$
Step 4 — Solve for the unknown (usually $v_f$ or $d$).
Worked Calculation
Substituting all values with units:
Ball of mass $m = 0.5\,\text{kg}$ dropped from $h = 10\,\text{m}$:
$$v_f = \sqrt{2gh} = \sqrt{2\times9.8\times10} = \sqrt{196} = 14\,\text{m/s}$$
Answer
$$\boxed{v_f = \sqrt{2g h}}$$
Physical Interpretation
A 14 m/s impact speed from a 10 m fall corresponds to about 50 km/h — enough to be seriously dangerous. This underscores why falling from height is hazardous. The conversion of gravitational PE to KE is 100% efficient in the absence of air resistance — every joule of lost PE appears as gained KE.
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