Grade 11 : Physics - Work/Power (Set 4)

Explanation: Work done is the scalar dot product of force and displacement vectors:
$W = \vec{F} \cdot \vec{d} = (3)(2) + (4)(-1) + (-2)(5)$
$W = 6 - 4 - 10 = -8\text{ J}$.

Explanation: Total work is the area under the $F-x$ curve:
Area 1 (rectangle, 0 to 4 m): $W_1 = 5 \times 4 = 20\text{ J}$
Area 2 (triangle, 4 to 8 m): $W_2 = \frac{1}{2} \times (8 - 4) \times 5 = 10\text{ J}$
Total Work $W = 20 + 10 = 30\text{ J}$.

Explanation: Work done equals change in kinetic energy:
$W = \Delta K = \frac{1}{2} m v^2 - 0 = \frac{1}{2} (1000) (20)^2 = 200,000\text{ J}$
Average power $P_{avg} = \frac{W}{\Delta t} = \frac{200,000}{5} = 40,000\text{ W} = 40\text{ kW}$.

Explanation: Mass of hanging part: $m' = \frac{M}{4} = 1kg$.
Hanging length: $h = \frac{L}{4} = 0.5m$.
Center of mass of hanging part is at depth $\frac{h}{2} = 0.25m$.
Work done $W = m' g \left(\frac{h}{2}\right) = 1 \times 10 \times 0.25 = 2.5\text{ J}$.

Explanation: Velocity at $t = 3\text{ s}$: $v = g t = 10 \times 3 = 30m/s$.
Gravitational force: $F = m g = 2 \times 10 = 20\text{ N}$.
Instantaneous power $P = F \cdot v = 20 \times 30 = 600\text{ W}$.

Explanation: Work done is the definite integral of force with respect to position:
$W = \int_{1}^{3} (3x^2 + 2x) dx = [x^3 + x^2]_{1}^{3}$
$W = (3^3 + 3^2) - (1^3 + 1^2) = (27 + 9) - (1 + 1) = 36 - 2 = 34\text{ J}$.

Explanation: Work by gravity: $W_g = m g L \sin\theta = 2 \times 10 \times 5 \times 0.6 = 60\text{ J}$.
Friction force: $f_k = \mu m g \cos\theta = 0.25 \times 2 \times 10 \times 0.8 = 4\text{ N}$.
Work by friction: $W_f = -f_k L = -4 \times 5 = -20\text{ J}$.
Net Work $W_{net} = W_g + W_f = 60 - 20 = 40\text{ J}$.

Explanation: Work done to lift: $W_1 = m g h = 1200 \times 10 \times 10 = 120,000\text{ J}$.
Work done for kinetic energy: $W_2 = \frac{1}{2} m v^2 = \frac{1}{2} \times 1200 \times 100 = 60,000\text{ J}$.
Total Work $W = 180,000\text{ J}$ in $t = 60\text{ s}$.
Power $P = \frac{180,000}{60} = 3000\text{ W} = 3\text{ kW}$.

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

Explanation: Calculate power $P = \frac{W}{t}$ for each machine:
M1: $5000 / 10 = 500\text{ W}$
M2: $7200 / 12 = 600\text{ W}$
M3: $4500 / 9 = 500\text{ W}$
M4: $6400 / 16 = 400\text{ W}$
Machine M2 delivers the highest power (600 W).

Explanation: The centripetal force is always perpendicular to the instantaneous displacement vector at every point along the circular path ($\theta = 90^\circ$).
Hence, $W = F_c \cdot d \cdot \cos 90^\circ = 0\text{ J}$.

Explanation: Since speed is constant, engine thrust equals resistive force: $F = 250\text{ N}$.
Power $P = F \cdot v = 250 \times 12 = 3000\text{ W} = 3.0\text{ kW}$.

Explanation: Calculate initial and final speeds:
At $t = 0\text{ s}$, $v_i = 3(0)^2 - 2(0) = 0m/s$
At $t = 2\text{ s}$, $v_f = 3(2)^2 - 2(2) = 12 - 4 = 8m/s$
By Work-Energy Theorem: $W = \frac{1}{2} m (v_f^2 - v_i^2) = \frac{1}{2} (2) (8^2 - 0) = 64\text{ J}$.

Explanation: Speed at $t = 3\text{ s}$ is $v = 2(3) = 6m/s$.
Centripetal force is perpendicular to velocity, delivering zero power.
Only tangential force delivers power: $P = F_t \cdot v = 5 \times 6 = 30\text{ W}$.

Explanation: Kinetic energy $K = \frac{p^2}{2m}$.
New momentum $p' = 1.5 p$.
New kinetic energy $K' = \frac{(1.5p)^2}{2m} = 2.25 K$.
Percentage increase = $\frac{K' - K}{K} \times 100% = 1.25 \times 100% = 125%$.

Explanation: Total upward force required $F = m g + f_{friction} = 1500(10) + 3000 = 18,000\text{ N}$.
Power $P = F \cdot v = 18,000 \times 3 = 54,000\text{ W} = 54\text{ kW}$.

Explanation: The profile forms a triangle with base $b = 8m$ and height $h = 20\text{ N}$.
Work done equals the area under the curve:
$W = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 20 = 80\text{ J}$.

Explanation: Work done by gravity depends only on vertical displacement:
$W_g = -m g \Delta h = -(0.5) \times 10 \times (10 - 0) = -50\text{ J}$.

Explanation: Work required: $W = \Delta K = \frac{1}{2} m v^2 = \frac{1}{2} \times 30 \times 10^2 = 1500\text{ J}$.
Time taken: $t = \frac{W}{P} = \frac{1500}{500} = 3seconds$.

Explanation: Useful output power $P_{out} = 0.80 \times 5000\text{ W} = 4000\text{ W}$.
Work output in 60 s: $W = 4000 \times 60 = 240,000\text{ J}$.
Since $W = m g h$, we have $240,000 = m \times 10 \times 20 = 200 m$.
Mass $m = \frac{240,000}{200} = 1200kg$.