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Machines - Energy and Efficiency

Grade 8IB

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

🔑Concepts

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Work is done when a force FF results in the displacement dd of an object in the direction of the force. It is measured in Joules (JJ).

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Power represents the rate at which work is performed or energy is converted. It is measured in Watts (WW), where 1W=1J/s1 W = 1 J/s.

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Mechanical Advantage (MAMA) is a measure of the force amplification achieved by using a tool or machine system. It is the ratio of the Load to the Effort.

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Velocity Ratio (VRVR) is the ratio of the distance moved by the effort to the distance moved by the load. In an ideal machine, MA=VRMA = VR.

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Efficiency (η\eta) indicates how much of the input energy is converted into useful output work. Due to friction and heat loss, efficiency is always less than 100%100\% in real machines.

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Energy Degradation occurs when energy is converted into non-useful forms, such as thermal energy due to friction between moving parts of a machine.

📐Formulae

W=F×dW = F \times d

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

MA=Load (F_L)Effort (F_E)MA = \frac{\text{Load (F\_{L})}}{\text{Effort (F\_{E})}}

VR=deffortdloadVR = \frac{d_{\text{effort}}}{d_{\text{load}}}

Efficiency (\eta)=(Work OutputWork Input)×100%\text{Efficiency (\eta)} = \left( \frac{\text{Work Output}}{\text{Work Input}} \right) \times 100\%

Efficiency=MAVR×100%\text{Efficiency} = \frac{MA}{VR} \times 100\%

💡Examples

Problem 1:

A pulley system is used to lift a load of 600 N600\text{ N} using an effort of 150 N150\text{ N}. If the effort moves 12 m12\text{ m} to lift the load 2.5 m2.5\text{ m}, calculate the Mechanical Advantage and the Velocity Ratio.

Solution:

MA=600150=4MA = \frac{600}{150} = 4 VR=122.5=4.8VR = \frac{12}{2.5} = 4.8

Explanation:

Mechanical Advantage is calculated by dividing the load by the effort. Velocity Ratio is calculated by dividing the distance the effort moves by the distance the load moves.

Problem 2:

A motor with an efficiency of 75%75\% is supplied with 2000 J2000\text{ J} of electrical energy. Calculate the useful work output and the energy wasted as heat.

Solution:

Useful Work=75100×2000=1500 J\text{Useful Work} = \frac{75}{100} \times 2000 = 1500\text{ J} 2000−1500500\begin{array}{r} 2000 \\ -1500 \\ \hline 500 \end{array} Energy Wasted=500 J\text{Energy Wasted} = 500\text{ J}

Explanation:

Useful work is found by multiplying the input energy by the efficiency percentage. The energy wasted is the difference between the input energy and the useful output energy.

Problem 3:

An electric hoist lifts a 250 N250\text{ N} weight to a height of 4 m4\text{ m} in 5 s5\text{ s}. Calculate the power output of the hoist.

Solution:

W=250×4=1000 JW = 250 \times 4 = 1000\text{ J} P=10005=200 WP = \frac{1000}{5} = 200\text{ W}

Explanation:

First, calculate the work done (F×dF \times d) to find the total energy transferred. Then, divide the work by the time taken to find the power.

Energy and Efficiency Grade 8 Notes & Examples | IB Science