Grade 11 : Physics - Work/Power (Set 3)

Explanation: Work done by a variable force is $W = \int_{x_1}^{x_2} F(x) dx = \int_{1}^{3} (3x^2 + 2x) dx = [x^3 + x^2]_1^3 = (3^3 + 3^2) - (1^3 + 1^2) = (27 + 9) - (1 + 1) = 36 - 2 = 34\text{ J}$.

Explanation: Displacement vector $\Delta \vec{r} = \vec{r}_2 - \vec{r}_1 = (5-1)\hat{i} + (7-2)\hat{j} = 4\hat{i} + 5\hat{j}m$. Work done $W = \vec{F} \cdot \Delta \vec{r} = (4)(4) + (3)(5) = 16 + 15 = 31\text{ J}$.

Explanation: Work done equals change in kinetic energy: $\Delta K = \frac{1}{2} m v^2 = \frac{1}{2} (1200) (20^2) = 240,000\text{ J}$. Average power $P_{avg} = \frac{\Delta K}{\Delta t} = \frac{240,000\text{ J}}{5\text{ s}} = 48,000\text{ W} = 48\text{ kW}$.

Explanation: Velocity at $t = 3\text{ s}$ is $v = gt = 9.8 \times 3 = 29.4m/s$. Force due to gravity is $F = mg = 2 \times 9.8 = 19.6\text{ N}$. Instantaneous power $P = F \cdot v = 19.6 \times 29.4 = 576.24\text{ W}$.

Explanation: Friction force $f_k = \mu_k mg = 0.25 \times 4 \times 10 = 10\text{ N}$. Work done by friction $W_f = -f_k d = -10d$. By work-energy theorem: $W_f = \Delta K = 0 - \frac{1}{2}(4)(6^2) = -72\text{ J}$. Thus $-10d = -72 \implies d = 7.2m$.

Explanation: Work done by spring force is $W_s = -\frac{1}{2}k(x_2^2 - x_1^2) = -\frac{1}{2}(400)(0.3^2 - 0.1^2) = -200(0.09 - 0.01) = -200(0.08) = -16\text{ J}$.

Explanation: Useful work output $W_{out} = mgh = 500 \times 10 \times 12 = 60,000\text{ J}$. Output power $P_{out} = \frac{60,000}{20} = 3000\text{ W}$. Input power $P_{in} = \frac{P_{out}}{\eta} = \frac{3000}{0.80} = 3750\text{ W} = 3.75\text{ kW}$.

Explanation: Acceleration $a = \frac{F}{m} = \frac{20}{5} = 4m/s^2$. Final velocity $v = at = 4 \times 4 = 16m/s$. By work-energy theorem, $W = \Delta K = \frac{1}{2}mv^2 - 0 = \frac{1}{2}(5)(16^2) = 2.5 \times 256 = 640\text{ J}$.

Explanation: Centripetal force acts perpendicular to the instantaneous velocity/displacement direction at every point ($ heta = 90^ op$). Thus, $W = F \cdot d \cos 90^ op = 0\text{ J}$.

Explanation: $W = \int_{0}^{2} (-20x + 4x^3) dx = [-10x^2 + x^4]_0^2 = (-10(4) + 16) - 0 = -40 + 16 = -24\text{ J}$.

Explanation: Power is $P = \frac{dW}{dt} \implies W = \int_{1}^{3} P(t) dt = \int_{1}^{3} (3t^2 + 2t) dt = [t^3 + t^2]_1^3 = (27 + 9) - (1 + 1) = 36 - 2 = 34\text{ J}$.

Explanation: Since speed is constant, pulling force equals parallel gravity component: $F = mg \sin 30^ op = 10 \times 10 \times 0.5 = 50\text{ N}$. Work done $W = F \cdot d = 50 \times 5 = 250\text{ J}$.

Explanation: Vertical height raised $h = L(1 - \cos 60^ op) = 1.5(1 - 0.5) = 0.75m$. Work done against gravity $W = mgh = 0.2 \times 10 \times 0.75 = 1.5\text{ J}$.

Explanation: Drag force at $v = 10m/s$ is $F_{drag} = 0.5 \times 10^2 = 50\text{ N}$. Power required is $P = F \cdot v = 50 \times 10 = 500\text{ W}$.

Explanation: $W_A = 100 \times 10 = 1000\text{ J}$, $W_B = 150 \times 6 = 900\text{ J}$, $W_C = 80 \times 12 = 960\text{ J}$. Total work = $1000 + 900 + 960 = 2860\text{ J}$.

Explanation: Statement I is true (definition of conservative force). Statement II is false because frictional force is non-conservative; work done by friction depends on path length and converts mechanical energy to heat.

Explanation: Total upward force needed $F = (m_{total})g + f = (800+200)(10) + 4000 = 10,000 + 4000 = 14,000\text{ N}$. Power output $P = F \cdot v = 14,000 \times 3 = 42,000\text{ W} = 42\text{ kW}$.

Explanation: By work-energy theorem, $F d = \frac{1}{2} m v^2 \implies d \propto v^2$. Doubling velocity increases stopping distance by $2^2 = 4$ times. Thus $d_2 = 4 \times 20m = 80m$.

Explanation: Mass of hanging part $m' = \frac{M}{4} = 1kg$. Length hanging $l' = \frac{L}{4} = 0.5m$. Center of gravity of hanging section lies at $h = \frac{l'}{2} = 0.25m$ below table surface. Work done $W = m' g h = 1 \times 10 \times 0.25 = 2.5\text{ J}$.

Explanation: Work equals area under $F-x$ curve. Triangular area ($x=0\text{ to }4m$): $\frac{1}{2} \times 4 \times 10 = 20\text{ J}$. Rectangular area ($x=4\text{ to }8m$): $4 \times 10 = 40\text{ J}$. Total Work = $20 + 40 = 60\text{ J}$.