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Physics: Motion and Mechanics - Work, Power, and Efficiency

Grade 8IB

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

🔑Concepts

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Work is defined as the product of the force (FF) applied to an object and the displacement (dd) of the object in the direction of the force. The SI unit of work is the Joule (JJ).

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Power is the rate at which work is done or energy is transferred. It is calculated by dividing the work done (WW) by the time taken (tt). The SI unit of power is the Watt (WW), where 1W=1J/s1 W = 1 J/s.

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Efficiency is a measure of how much of the input energy to a system is converted into useful output energy. It is usually expressed as a percentage.

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Energy is the capacity to do work. Work done on an object results in a change in its energy, such as Kinetic Energy (KEKE) or Gravitational Potential Energy (GPEGPE).

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According to the Law of Conservation of Energy, energy cannot be created or destroyed, only transformed from one form to another. In real-world machines, some energy is always 'lost' as heat due to friction, making efficiency less than 100%100\%.

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Gravitational Potential Energy (GPEGPE) is calculated using the mass (mm), gravitational acceleration (g≈9.8m/s2g \approx 9.8 m/s^2), and height (hh).

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Kinetic Energy (KEKE) depends on the mass (mm) and the square of the velocity (vv) of the object.

📐Formulae

W=F×dW = F \times d

P=WtP = \frac{W}{t}

Efficiency=(Useful Energy OutputTotal Energy Input)×100%\text{Efficiency} = \left( \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \right) \times 100\%

GPE=m×g×hGPE = m \times g \times h

KE=12mv2KE = \frac{1}{2} m v^2

1J=1N⋅m1 J = 1 N \cdot m

💡Examples

Problem 1:

A person exerts a force of 200N200 N to push a crate across a floor for a distance of 5m5 m. Calculate the work done.

Solution:

W=F×dW = F \times d W=200N×5mW = 200 N \times 5 m W=1000JW = 1000 J

Explanation:

Work is the product of force and displacement. Since the force and displacement are in the same direction, we multiply them directly to get 10001000 Joules.

Problem 2:

A crane lifts a load of 5000J5000 J in 20s20 s. What is the power generated by the crane?

Solution:

P=WtP = \frac{W}{t} P=5000J20sP = \frac{5000 J}{20 s} P=250WP = 250 W

Explanation:

Power is work divided by time. Dividing the 5000J5000 J of work by the 2020 seconds it took gives a power rating of 250250 Watts.

Problem 3:

An electric motor takes in 800J800 J of electrical energy. It performs 600J600 J of useful work. Calculate the efficiency of the motor and the energy lost to the surroundings.

Solution:

Efficiency: Efficiency=600J800J×100%\text{Efficiency} = \frac{600 J}{800 J} \times 100\% Efficiency=0.75×100%=75%\text{Efficiency} = 0.75 \times 100\% = 75\% Energy lost: 800−600200\begin{array}{r} 800 \\ - 600 \\ \hline 200 \end{array} Energy lost = 200J200 J.

Explanation:

Efficiency is the ratio of useful output to total input. The motor is 75%75\% efficient. The remaining 200J200 J of energy is typically dissipated as heat and sound.