WORK AND ENERGY (Continued..)

Core Principles of Work & Energy

• Mechanical Energy ($E_{mech}$)

  • Sum of translational Kinetic Energy ($K$) and Potential Energy ($U$): $E_{mech}=K+U$.

• Work Done by a Force ($W$)

  • Definition: $W = \int \vec F \cdot d\vec r = F\,d\cos\theta$ (for constant $\vec F$).

  • Positive work adds energy to a system; negative work removes energy.

• Work–Energy Theorem

  • $\Delta K = W<em>{net}$, where $W</em>{net}$ is the algebraic sum of the work done by all forces.

• Conservative vs. Non-conservative Forces

  • Conservative (e.g.
    gravity, ideal spring): path–independent work, associated potential $U$, closed-loop work $=0$.

  • Non-conservative (e.g.
    kinetic friction, air drag): path-dependent; their work changes the system’s total mechanical energy.

• Energy Accounting Equation

  • $\boxed{\Delta E<em>{mech}=W</em>{nc}}$ (change in mechanical energy equals work done by non-conservative forces).

Conceptual Question Set (Pages 2 – 4)

• Page 2 – Positive work by a net external non-conservative force

• Given: W_{nc}>0.

• From $\Delta E<em>{mech}=W</em>{nc}$ ⇒ $E_{mech}$ increases.

• Potential and kinetic may trade off internally, so only total can be guaranteed.

• Correct conclusion → E. Total mechanical energy increases.

• Page 3 – Two forces $F1$ & $F2$ increase speed (so \Delta K>0)

• We test each option against W{net}=W{1}+W_{2}>0.

• A: $+0$ → >0 ✔

• B: $0+$ → >0 ✔

• C: $++$ → definitely >0 ✔

• D: $--$ → sum <0 ⇒ speed would drop ✖ (NOT possible)

• E: $+ -$ could still net positive if W1>|W2| ✔

• Answer → D. Both works negative is impossible if speed increases.

• Page 4 – Ferris-wheel rider, one full revolution

• Gravity is conservative; closed path ⇒ work$=0$.

• Answer → C. Net gravitational work is zero regardless of speed or diameter.

Quantitative Problem 1 (Page 5) – Skydiver With Open Parachute

• Data: $m=92.0\,\text{kg},\; \Delta h = 325\,\text{m (down)},\; v = \text{constant}$.

• $\Delta K =0$ (speed constant).

• Change in potential: $\Delta U = -mg\Delta h = -(92)(9.8)(325) = -2.93\times10^{5}\,\text{J}$.

• $\Delta E_{mech}=\Delta K+\Delta U = -2.93\times10^{5}\,\text{J}$.

• From $W<em>{nc}=\Delta E</em>{mech}$ → work by air resistance = $-2.93\times10^{5}\,\text{J}$ (removes energy).

• Answer → A. $-2.93\times10^{5}\,\text{J}$.

Quantitative Problem 2 (Page 6) – Projectile With Air Resistance

(a) Maximum height w/o air drag

• Using energy: $K<em>i + U</em>i = K<em>{top}+U</em>{top}$, with $K_{top}=0$.

• $\frac12 mv<em>0^2 = mg h</em>{ideal}$ ⇒ $h<em>{ideal}= \dfrac{v</em>0^2}{2g}= \dfrac{18^2}{2(9.8)} \approx 16.5\,\text{m}$.

(b) Average resisting force when only $11.8\,\text{m}$ reached

• Actual $\Delta U = mg h_{real} = (0.750)(9.8)(11.8)= 86.7\,\text{J}$.

• Initial kinetic energy $= \dfrac12(0.750)(18)^2 = 121\,\text{J}$.

• Mechanical‐energy change $\Delta E<em>{mech}=U</em>{top}-K_i = 86.7-121 = -34.3\,\text{J}$ (loss).

• Work by air $=\Delta E_{mech}= -34.3\,\text{J}$.

• If average drag force $F_d$ opposes upward displacement $d=11.8\,\text{m}$:

  • $W = -F_d\,d$ (negative sign built-in since force opposite displacement).

  • $F_d = |W|/d = 34.3/11.8 \approx 2.9\,\text{N}$.

• Magnitude of average air-resistance force ≈ $2.9\,\text{N}$.

Connections, Significance & Real-World Context

• Ferris-wheel & projectile examples illustrate that gravity is conservative: energy bookkeeping around loops or between heights is straightforward.

• Skydiver problem demonstrates how non-conservative forces (drag) convert mechanical energy into thermal energy, permitting terminal (constant) velocity.

• Conceptual multiple-choice tasks test one’s ability to apply $\Delta E<em>{mech}=W</em>{nc}$ and the work–energy theorem without numeric crunching—essential exam skills.

• Practical implications:

  • Engineers size parachutes by equating drag work to gravitational energy loss for acceptable descent rates.

  • Amusement-ride safety analyses rely on knowing that gravity alone cannot change a rider’s mechanical energy over closed cycles, so motors/ brakes supply necessary non-conservative work.

• Ethical note: misestimating non-conservative forces (e.g.
  underrating air drag) can lead to catastrophic over-speeds or insufficient safety margins.