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{net}, where W{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{mech}=W{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{mech}=W{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{nc}=\Delta E{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: Ki + Ui = K{top}+U{top}, with K_{top}=0.
\frac12 mv0^2 = mg h{ideal} ⇒ h{ideal}= \dfrac{v0^2}{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{mech}=U{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{mech}=W{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.