Problem 3.149 — Mutual inductance energy

Problem Statement

Solve the work-energy problem: Problem 3.149 — Mutual inductance energy $k = M/\sqrt{L_1L_2} = 3/$ This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and

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.