Work, Energy and Power

(a) Work done by man lifting bucket:
Force is upward, displacement is upward. Angle $\theta = 0^\circ$.
$W = Fs \cos 0^\circ$. POSITIVE.

(b) Work done by gravity in the above case:
Gravity acts downward, displacement is upward. $\theta = 180^\circ$.
$W = Fs \cos 180^\circ$. NEGATIVE.

(c) Work done by friction on sliding body:
Friction opposes motion. $\theta = 180^\circ$. NEGATIVE.

(d) Applied force on rough horizontal plane (uniform velocity):
Force is in direction of motion to overcome friction. $\theta = 0^\circ$. POSITIVE.

(e) Resistive force of air on pendulum:
Resistive force always opposes motion. $\theta = 180^\circ$. NEGATIVE.

Given: $m=2 \text{ kg}$, $u=0$, $F_{app}=7 \text{ N}$, $\mu_k=0.1$.
Friction $f_k = \mu_k N = \mu_k mg = 0.1 \times 2 \times 9.8 = 1.96 \text{ N}$.
Net Force $F_{net} = 7 – 1.96 = 5.04 \text{ N}$.
Acceleration $a = F_{net}/m = 5.04/2 = 2.52 \text{ m/s}^2$.
Displacement in $t=10 \text{ s}$: $s = ut + \frac{1}{2}at^2 = 0 + 0.5(2.52)(100) = 126 \text{ m}$.

(a) Work by applied force: $W_F = 7 \times 126 = \mathbf{882 \text{ J}}$.

(b) Work by friction: $W_f = -1.96 \times 126 = \mathbf{-246.96 \text{ J}}$.

(c) Work by net force: $W_{net} = 5.04 \times 126 = \mathbf{635.04 \text{ J}}$.

(d) Change in KE: By Work-Energy Theorem, $\Delta K = W_{net} = \mathbf{635.04 \text{ J}}$.

Concept: A particle cannot exist where Total Energy $E < V(x)$ because Kinetic Energy $K = E - V(x)$ would be negative (impossible).

• Fig (a): $x > a$. Since $V(x) > E$ for $x < a$ and $x > b$? No, look at the graph line. The particle is confined where the solid line is below the cross.
• Analysis: Identify regions where the potential curve $V(x)$ goes above the total energy line $E$. These are forbidden.

(a) Rocket casing burns: Heat energy comes from the Rocket’s Mechanical Energy (Kinetic + Potential) working against air resistance.

(b) Comets work done zero: Gravity is a Conservative Force. Work done in a closed path (complete orbit) is always zero.

(c) Satellite speed increases: Friction reduces total energy $E = -GMm/2r$. As $E$ becomes more negative (decreases), radius $r$ decreases. Since orbital speed $v \propto 1/\sqrt{r}$, as $r$ decreases, $v$ increases. (Potential energy decreases faster than Kinetic energy increases).

(d) Man pulling vs carrying:
Carrying: Displacement is horizontal, Force (against gravity) is vertical. $W = 0$ (Physics definition).
Pulling: [Image of pulley system] Force is applied along the rope to lift mass against gravity (vertical displacement). $W = mgh$.
Case (ii) involves greater work against gravity.

(a) Elastic collision, momentum and energy conserved: FALSE. Total Energy is conserved, but Energy usually implies Kinetic Energy. Kinetic Energy is conserved only after the collision, not necessarily during. Momentum is conserved.

(b) Total energy always conserved: TRUE. (Including heat, sound, etc., for isolated systems).

(c) Work in closed loop is zero for every force: FALSE. Only true for Conservative forces (like Gravity). False for Non-conservative forces (like Friction).

(d) Inelastic collision, final KE < initial KE: TRUE. Energy is lost as heat/deformation.

Q5.9 Body starting from rest, constant acceleration $a$:
$v = at$. Force $F = ma$.
Power $P = F \cdot v = (ma)(at) = ma^2 t$.
So, $P \propto t$. Answer: (ii)

Q5.10 Constant Power $P$:
$P = \text{const} \Rightarrow \frac{dK}{dt} = P \Rightarrow K = Pt$.
$\frac{1}{2}mv^2 = Pt \Rightarrow v = \sqrt{\frac{2P}{m}} t^{1/2}$.
$\frac{dx}{dt} \propto t^{1/2}$. Integrating both sides:
$x \propto t^{3/2}$. Answer: (iii)

Kinetic Energy $K = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2K}{m}}$.
$K_e = 10 \text{ keV}$, $K_p = 100 \text{ keV}$. $m_p \approx 1836 m_e$.

Ratio $\frac{v_e}{v_p} = \sqrt{\frac{K_e}{m_e} \times \frac{m_p}{K_p}} = \sqrt{\frac{10}{1} \times \frac{1836}{100}} = \sqrt{183.6} \approx 13.5$.
The electron is faster.

Given: $r=2 \text{ mm}$, $h=500 \text{ m}$.
Mass $m = \text{Vol} \times \rho = \frac{4}{3}\pi (2\times 10^{-3})^3 \times 1000 \approx 3.35 \times 10^{-5} \text{ kg}$.

Work by Gravity (First Half, $h=250$ m):
$W_g = mgh = 3.35 \times 10^{-5} \times 9.8 \times 250 \approx \mathbf{0.082 \text{ J}}$.

Work by Resistive Force (Total Journey):
Change in KE $\Delta K = \frac{1}{2}mv^2 – 0 = 0.5 \times 3.35 \times 10^{-5} \times (10)^2 = 1.675 \times 10^{-3} \text{ J}$.
Total Work $W_{net} = W_g + W_r = \Delta K$.
$W_g (\text{Total}) = mgh_{total} \approx 0.164 \text{ J}$.
$W_r = \Delta K – W_g = 0.001675 – 0.164 \approx \mathbf{-0.162 \text{ J}}$.

Volume $V = 30 \text{ m}^3 \Rightarrow \text{Mass } m = 30,000 \text{ kg}$.
Height $h = 40 \text{ m}$. Time $t = 15 \text{ min} = 900 \text{ s}$.
Efficiency $\eta = 30\% = 0.3$.

Output Power $P_{out} = \frac{mgh}{t} = \frac{30000 \times 9.8 \times 40}{900} = 13066.7 \text{ W}$.
Input Power $P_{in} = \frac{P_{out}}{\eta} = \frac{13066.7}{0.3} \approx 43555 \text{ W} = \mathbf{43.6 \text{ kW}}$.

Concept: In an elastic head-on collision between identical masses, velocities are exchanged.
Ball 1 ($v$) hits Ball 2 (rest). Ball 1 stops, Ball 2 moves with $v$.
Ball 2 hits Ball 3 (rest). Ball 2 stops, Ball 3 moves with $v$.
Result: The incident ball stops, and exactly one ball from the other end moves out with speed $v$. (Fig ii).

(a) Mass passing area A:
Volume flux = $Av$. Mass flux = $\rho A v$. Mass in time $t$: $M = \rho A v t$.

(b) Kinetic Energy:
$K = \frac{1}{2} M v^2 = \frac{1}{2} (\rho A v t) v^2 = \frac{1}{2} \rho A v^3 t$.

(c) Electrical Power:
$P_{wind} = \frac{K}{t} = \frac{1}{2} \rho A v^3$.
$v = 36 \text{ km/h} = 10 \text{ m/s}$. $A=30$. $\rho=1.2$.
$P_{wind} = 0.5 \times 1.2 \times 30 \times (10)^3 = 18,000 \text{ W}$.
Electric Power (25% eff) $= 0.25 \times 18000 = \mathbf{4500 \text{ W}} = \mathbf{4.5 \text{ kW}}$.