Grade 11 : Physics - Work/Power (Set 2)

Explanation: Work done is calculated using $W = F \cdot d \cdot \cos\theta$. Here, $\theta = 0^\circ$, so $W = 50\text{ N} \times 10m \times 1 = 500\text{ J}$.

Explanation: The standard SI unit of work is the Joule ($\text{J}$), which is equivalent to one Newton-meter ($1\text{ N}\cdotm = 1kg\cdotm^2/\text{s}^2$).

Explanation: The gravitational force acts radially inward, perpendicular to the circular motion displacement ($\theta = 90^\circ$). Thus, $W = F d \cos 90^\circ = 0\text{ J}$.

Explanation: Power is defined as the rate of doing work: $P = \frac{W}{t} = \frac{600\text{ J}}{12\text{ s}} = 50\text{ W}$.

Explanation: Work done against gravity equals gravitational potential energy gained: $W = mgh = 4 \times 10 \times 5 = 200\text{ J}$.

Explanation: By standard engineering definition, $1\text{ horsepower (hp)} \approx 746\text{ Watts (W)}$.

Explanation: Friction acts opposite to displacement ($\theta = 180^\circ$). $W = F d \cos 180^\circ = 5 \times 5 \times (-1) = -25\text{ J}$.

Explanation: Power delivered by a force moving at speed $v$ is given by $P = F \cdot v = 100\text{ N} \times 3m/s = 300\text{ W}$.

Explanation: Mathematically, work done is the integral of force with respect to displacement ($W = \int F dx$), which corresponds to the area under an $F-x$ curve.

Explanation: Work done to compress a spring is $W = \frac{1}{2} k x^2 = \frac{1}{2} \times 200 \times (0.2)^2 = 100 \times 0.04 = 4\text{ J}$.

Explanation: $1\text{ kWh} = 1000\text{ W} \times 3600\text{ s} = 3,600,000\text{ J} = 3.6 \times 10^6\text{ J}$.

Explanation: The vertical force applied is perpendicular to the horizontal displacement ($\theta = 90^\circ$), so $W = F d \cos 90^\circ = 0\text{ J}$.

Explanation: $W = mgh = 500 \times 10 \times 10 = 50,000\text{ J}$. Power $P = \frac{W}{t} = \frac{50,000\text{ J}}{10\text{ s}} = 5000\text{ W} = 5\text{ kW}$.

Explanation: Power is Work divided by time: $[P] = \frac{[M L^2 T^{-2}]}{[T]} = [M L^2 T^{-3}]$.

Explanation: Work is Force $\times$ Distance $= [M L T^{-2}] \times [L] = [M L^2 T^{-2}]$.

Explanation: Using $W = F d \cos\theta = 40 \times 6 \times \cos(60^\circ) = 240 \times 0.5 = 120\text{ J}$.

Explanation: Work done equals change in kinetic energy: $W = \frac{1}{2} m v^2 = \frac{1}{2} \times 2 \times 10^2 = 100\text{ J}$. Average power $P = \frac{W}{t} = \frac{100}{5} = 20\text{ W}$.

Explanation: Work is a scalar quantity resulting from the scalar (dot) product of force and displacement vectors: $W = \vec{F} \cdot \vec{d}$.

Explanation: Efficiency is $\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100% = \frac{800}{1000} \times 100% = 80%$.

Explanation: Work $W = \vec{F} \cdot \vec{d} = (3)(2) + (4)(5) = 6 + 20 = 26\text{ J}$.