Power

Practice Questions:

Question 11: easy

Water falls from a height of \(60\text{ m}\) at the rate of \(15\text{ kg/s}\) to operate a turbine. The losses due to frictional force are \(10%\) of the input energy. How much power is generated by the turbine? (\(g = 10\text{ m/s}^2\))

1. 7.0 kW

2. 10.2 kW

3. 8.1 kW

4. 12.3 kW

View Answer

The total input power is \(P_{\text{in}} = \frac{dm}{dt} gh = 15 \times 10 \times 60 = 9000\text{ W} = 9\text{ kW}\). Given that \(10%\) of energy is lost to friction, the efficiency is \(90%\), so the power generated is \(P_{\text{out}} = 0.9 \times 9\text{ kW} = 8.1\text{ kW}\).

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Question 12: easy

A particle moves with a velocity \((5\hat{i} – 3\hat{j} + 6\hat{k}) \text{ms}^{-1}\) horizontally under the action of constant force \((10\hat{i} + 10\hat{j} + 20\hat{k}) \text{N}\). The instantaneous power supplied to the particle is :

1. 100 W

2. 140 W

3. 200 W

4. Zero

View Answer

Instantaneous power is calculated using the dot product of force and velocity: \(P = \vec{F} \cdot \vec{v} = (10)(5) + (10)(-3) + (20)(6) = 50 - 30 + 120 = 140 \text{W}\).

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Question 13: easy

The power of a water pump is \(100 \text{kW}\). If \(g = 10 \text{m/s}^2\), then the amount of water it can raise in \(2 \text{s}\) to a height of \(20 \text{m}\) is

1. \(200 \text{kg}\)

2. \(300 \text{kg}\)

3. \(400 \text{kg}\)

4. \(1000 \text{kg}\)

View Answer

Power \(P = \frac{mgh}{t} ⇒ 100 \times 10^3 = \frac{m \times 10 \times 20}{2}\). This simplifies to \(100 \times 10^3 = 100 m ⇒ m = 1000 \text{kg}\).

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Question 14: moderate

The position of a particle at any time \( t \) is given by \( x = t^2 + 1 \), where \( x \) is in m and \( t \) is in s. If a constant force of 8 N is acting on the particle, then the instantaneous power of the force at \( t = 1\text{ s} \) will be

1. 8 W

2. 16 W

3. 32 W

4. 19 W

View Answer

The velocity of the particle is \( v = \frac{dx}{dt} = 2t \). At \( t = 1\text{ s} \), \( v = 2\text{ m/s} \). The instantaneous power is \( P = F \cdot v = 8\text{ N} \times 2\text{ m/s} = 16\text{ W} \).

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