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Machines - Work

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 on an object and the displacement ss of the object in the direction of the force. The SI unit for work is the Joule (JJ), where 1 J=1 Nm1\text{ J} = 1\text{ Nm}.

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Power (PP) is the rate at which work is done. Its SI unit is the Watt (WW), representing one Joule per second (1 W=1 J/s1\text{ W} = 1\text{ J/s}).

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A Machine is a device that can change the direction of force or multiply the effort applied to overcome a load. It helps in doing work more easily or quickly.

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Mechanical Advantage (MAMA) is the ratio of the load (LL) to the effort (EE). It indicates how many times a machine multiplies the input force.

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Velocity Ratio (VRVR) is the ratio of the distance moved by the effort (dEd_E) to the distance moved by the load (dLd_L). It depends solely on the geometry of the machine.

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Efficiency (η\eta) is the ratio of useful work output to the work input. For an ideal machine, efficiency is 100%100\%, but in real machines, it is always less than 100%100\% due to friction and the weight of moving parts.

📐Formulae

W=F×sW = F \times s

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

MA=Load(L)Effort(E)MA = \frac{Load (L)}{Effort (E)}

VR=dEdLVR = \frac{d_E}{d_L}

η=Work OutputWork Input×100%\eta = \frac{Work\,Output}{Work\,Input} \times 100\%

η=MAVR×100%\eta = \frac{MA}{VR} \times 100\%

💡Examples

Problem 1:

A crane lifts a load of 2500 N2500\text{ N} to a height of 20 m20\text{ m} in 50 seconds50\text{ seconds}. Calculate the work done and the power of the crane.

Solution:

Work done W=F×s=2500×20=50000 JW = F \times s = 2500 \times 20 = 50000\text{ J}. Power P=Wt=5000050=1000 WP = \frac{W}{t} = \frac{50000}{50} = 1000\text{ W}.

Explanation:

We first use the work formula to find the energy transferred and then divide by time to find the rate of work (Power).

Problem 2:

A simple machine with a velocity ratio of 55 is used to lift a load of 600 N600\text{ N} by applying an effort of 150 N150\text{ N}. Calculate the Mechanical Advantage and Efficiency.

Solution:

MA=LE=600150=4MA = \frac{L}{E} = \frac{600}{150} = 4 η=MAVR×100%=45×100=80%\eta = \frac{MA}{VR} \times 100\% = \frac{4}{5} \times 100 = 80\%

Explanation:

The MAMA is calculated by dividing load by effort. The efficiency is the ratio of MAMA to VRVR, showing that 20%20\% of the energy is lost, likely to friction.

Problem 3:

Calculate the total work done if a worker performs two tasks: Task A requires 4580 J4580\text{ J} and Task B requires 3245 J3245\text{ J}.

Solution:

4580+32457825\begin{array}{r} 4580 \\ +3245 \\ \hline 7825 \end{array} Total Work =7825 J= 7825\text{ J}.

Explanation:

The total work done is the sum of the work done in individual tasks, calculated using vertical addition.

Work Grade 8 Notes & Examples | IB Science