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Work, Energy and Simple Machines - Simple Machines

Grade 9CBSE

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

🔑Concepts

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A Simple Machine is a device that allows work to be done more easily by changing the magnitude or direction of the applied force.

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Mechanical Advantage (MAMA) is the ratio of the Load (LL) to the Effort (EE). It indicates how many times the machine multiplies the effort: MA=LEMA = \frac{L}{E}.

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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): VR=dEdLVR = \frac{d_E}{d_L}.

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Efficiency (η\eta) is the ratio of useful work output to the work input. For an ideal machine, η=1\eta = 1 (or 100%100\%).

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In a real machine, Efficiency is always less than 11 due to friction and the weight of moving parts, meaning MA<VRMA < VR.

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Levers are classified into three types: Class I (Fulcrum in between, e.g., seesaw), Class II (Load in between, e.g., wheelbarrow), and Class III (Effort in between, e.g., tweezers).

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An Inclined Plane reduces the effort required to lift a load by increasing the distance over which the force is applied.

📐Formulae

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

VR=Distance moved by Effort (dE)Distance moved by Load (dL)VR = \frac{\text{Distance moved by Effort } (d_E)}{\text{Distance moved by Load } (d_L)}

η=Work OutputWork Input=MAVR\eta = \frac{\text{Work Output}}{\text{Work Input}} = \frac{MA}{VR}

Work Input=E×dE\text{Work Input} = E \times d_E

Work Output=L×dL\text{Work Output} = L \times d_L

\text{Efficiency (in %)} = \left( \frac{MA}{VR} \right) \times 100

💡Examples

Problem 1:

A machine is used to lift a load of 500 N500 \text{ N} by applying an effort of 100 N100 \text{ N}. If the effort moves 2.5 m2.5 \text{ m} while the load moves 0.4 m0.4 \text{ m}, calculate the Mechanical Advantage and Efficiency.

Solution:

MA=LE=500 N100 N=5MA = \frac{L}{E} = \frac{500 \text{ N}}{100 \text{ N}} = 5 VR=dEdL=2.5 m0.4 m=6.25VR = \frac{d_E}{d_L} = \frac{2.5 \text{ m}}{0.4 \text{ m}} = 6.25 η=MAVR=56.25=0.8\eta = \frac{MA}{VR} = \frac{5}{6.25} = 0.8 \eta \text{ (in %)} = 0.8 \times 100 = 80\%

Explanation:

The machine multiplies the force by 55 times (MAMA), but the effort moves 6.256.25 times more distance than the load (VRVR). The efficiency is 80%80\% because some energy is lost to friction.

Problem 2:

Calculate the energy lost as heat if the total work input into a machine is 4500 J4500 \text{ J} and the useful work output is 3825 J3825 \text{ J}.

Solution:

To find the energy lost, we subtract the output work from the input work: 4500−3825675\begin{array}{r} 4500 \\ -3825 \\ \hline 675 \end{array} The energy lost is 675 J675 \text{ J}.

Explanation:

Energy loss is the difference between the total energy provided to the system and the energy successfully converted into useful work.