Grade 11 : Physics - Work/Power (Set 1)

Explanation: Work done is calculated using $W = F \cdot d \cdot \cos\theta$. Here, $F = 10\text{ N}$, $d = 5m$, and $\theta = 0^\circ$. Thus, $W = 10 \times 5 \times \cos(0^\circ) = 50\text{ J}$.

Explanation: Work done is $W = F \cdot d \cdot \cos\theta$. When force and displacement are perpendicular, $\theta = 90^\circ$, so $\cos(90^\circ) = 0$, resulting in $W = 0\text{ J}$.

Explanation: Power is defined as the rate of doing work ($P = \frac{W}{t}$). The SI unit of power is Joules per second ($\text{J/s}$), which is defined as the Watt (W).

Explanation: Power is calculated using $P = \frac{W}{t} = \frac{200\text{ J}}{10\text{ s}} = 20\text{ W}$.

Explanation: Work done against gravity is given by $W = mgh = 2 \times 9.8 \times 3 = 58.8\text{ J}$.

Explanation: Frictional force acts opposite to the displacement direction ($\theta = 180^\circ$). Since $\cos(180^\circ) = -1$, the work done by friction is negative.

Explanation: Since $P = \frac{dW}{dt} = \frac{\vec{F} \cdot d\vec{r}}{dt} = \vec{F} \cdot \vec{v}$, instantaneous power is the scalar dot product of force and velocity.

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

Explanation: $W = F \cdot d \cdot \cos(60^\circ) = 25 \times 8 \times 0.5 = 100\text{ J}$.

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

Explanation: Centripetal force acts towards the center while displacement is tangential. Since the angle between force and displacement is always $90^\circ$, work done is zero.

Explanation: $\text{Power} = \frac{\text{Work}}{\text{Time}} = \frac{[M L^2 T^{-2}]}{[T]} = [M L^2 T^{-3}]$.

Explanation: By the Work-Energy Theorem, $W_{\text{net}} = \Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) = \frac{1}{2}(4)(6^2 - 2^2) = 2(36 - 4) = 64\text{ J}$.

Explanation: Convert time to seconds: $1minutes = 60\text{ s}$. Power $P = \frac{W}{t} = \frac{3600\text{ J}}{60\text{ s}} = 60\text{ W}$.

Explanation: Kinetic energy is proportional to the square of velocity ($KE = \frac{1}{2}mv^2$). Doubling speed ($2v$) increases kinetic energy by $2^2 = 4$ times.

Explanation: Work done by spring restoring force is $W_{\text{spring}} = -\frac{1}{2}kx^2 = -\frac{1}{2}(100)(0.2)^2 = -0.5 \times 100 \times 0.04 = -2.0\text{ J}$.

Explanation: Displacement $d = 0$. Since work requires displacement ($W = F \cdot d \cos\theta$), zero displacement means zero work is done on the bag.

Explanation: $\text{Efficiency} = \left(\frac{P_{\text{out}}}{P_{\text{in}}}\right) \times 100% = \left(\frac{400}{500}\right) \times 100% = 80%$.

Explanation: Power for constant speed vertical motion is $P = F \cdot v = 1500\text{ N} \times 2m/s = 3000\text{ W} = 3\text{ kW}$.

Explanation: $W = \int_{x_1}^{x_2} F(x) dx = \int_{0}^{2} 3x^2 dx = [x^3]_0^2 = 2^3 - 0 = 8\text{ J}$.