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Work, Energy, and Power - Power

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Power is defined as the time rate at which work is done or energy is transferred. It is a scalar quantity with dimensions [ML2T−3][ML^2T^{-3}].

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Average power is the ratio of the total work done WW to the total time interval tt taken to do that work: Pavg=WtP_{avg} = \frac{W}{t}.

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Instantaneous power is the power delivered at a specific instant of time, defined as the derivative of work with respect to time: P=dWdtP = \frac{dW}{dt}.

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Power can be expressed in terms of force F⃗\vec{F} and velocity v⃗\vec{v} as the dot product: P=F⃗⋅v⃗=Fvcos⁡θP = \vec{F} \cdot \vec{v} = Fv \cos \theta, where θ\theta is the angle between force and velocity vectors.

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The SI unit of power is the Watt (WW), where 1 W=1 J/s1 \text{ W} = 1 \text{ J/s}. Other common units include Horsepower (hp), where 1 hp=746 W1 \text{ hp} = 746 \text{ W}.

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The efficiency η\eta of a device is the ratio of useful output power to the total input power, often expressed as a percentage: η=(PoutPin)×100%\eta = \left( \frac{P_{out}}{P_{in}} \right) \times 100\%.

📐Formulae

Pavg=WtP_{avg} = \frac{W}{t}

P=dWdtP = \frac{dW}{dt}

P=F⃗⋅v⃗P = \vec{F} \cdot \vec{v}

P=Fvcos⁡θP = Fv \cos \theta

1 hp=746 W1 \text{ hp} = 746 \text{ W}

η=PoutputPinput×100%\eta = \frac{P_{output}}{P_{input}} \times 100\%

💡Examples

Problem 1:

An elevator can carry a maximum load of 1800 kg1800 \text{ kg} (elevator + passengers) and is moving up with a constant speed of 2 m/s2 \text{ m/s}. The frictional force opposing the motion is 4000 N4000 \text{ N}. Determine the minimum power delivered by the motor to the elevator in Watts and Horsepower. (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

The downward forces acting on the elevator are the weight W=mgW = mg and the frictional force ff. Total downward force F=mg+fF = mg + f. F=(1800×10)+4000=18000+4000=22000 NF = (1800 \times 10) + 4000 = 18000 + 4000 = 22000 \text{ N} To move with constant speed, the motor must apply an upward force equal to 22000 N22000 \text{ N}. Power is given by P=F⋅vP = F \cdot v. P=22000×2=44000 WP = 22000 \times 2 = 44000 \text{ W} To convert to Horsepower: Php=44000746≈58.98 hpP_{hp} = \frac{44000}{746} \approx 58.98 \text{ hp}

Explanation:

Since the speed is constant, the net force is zero. The motor must counteract both gravity and friction. The power is the product of this total resistive force and the constant velocity.

Problem 2:

A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m330 \text{ m}^3 in 15 min15 \text{ min}. If the tank is 40 m40 \text{ m} above the ground, and the efficiency of the pump is 30%30\%, how much electric power is consumed by the pump? (Density of water ρ=103 kg/m3\rho = 10^3 \text{ kg/m}^3, g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

Mass of water m=Volume×Density=30×103=30000 kgm = \text{Volume} \times \text{Density} = 30 \times 10^3 = 30000 \text{ kg}. Work done to lift water W=mgh=30000×10×40=1.2×107 JW = mgh = 30000 \times 10 \times 40 = 1.2 \times 10^7 \text{ J}. Output power Pout=Wt=1.2×10715×60=1.2×107900≈13333.33 WP_{out} = \frac{W}{t} = \frac{1.2 \times 10^7}{15 \times 60} = \frac{1.2 \times 10^7}{900} \approx 13333.33 \text{ W}. Given efficiency η=30%\eta = 30\%, Pin=Poutη×100P_{in} = \frac{P_{out}}{\eta} \times 100. Pin=13333.330.3≈44444.44 WP_{in} = \frac{13333.33}{0.3} \approx 44444.44 \text{ W}

Explanation:

First, calculate the potential energy gained by the water (Work). Then find the useful output power by dividing work by time. Finally, account for efficiency to find the total electrical power consumed (Input Power).