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Force, Work and Energy - Simple Machines (Lever, Pulley, Inclined Plane, Screw, Wedge, Wheel and Axle)

Grade 5ICSE

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

🔑Concepts

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A Simple Machine is a device that makes work easier by changing the direction of force, increasing the speed, or decreasing the effort required.

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Work is done only when a Force applied on an object causes it to move through a certain distance in the direction of the force. It is represented as W=FimesdW = F imes d.

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A Lever consists of a rigid bar that moves around a fixed point called the Fulcrum (FF). It involves an Effort (EE) used to move a Load (LL).

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

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A Pulley consists of a grooved wheel and a rope. A Fixed Pulley changes the direction of force, while a Movable Pulley reduces the amount of effort needed.

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An Inclined Plane is a slanting surface (ramp) used to lift heavy loads to a certain height with less effort over a longer distance.

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A Screw is essentially an inclined plane wrapped around a cylinder in a spiral. The spiral ridges are called threads.

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A Wedge consists of two inclined planes joined back-to-back, used to split or cut objects (e.g., an axe or a knife).

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The Wheel and Axle consists of a larger wheel attached to a smaller rod called an axle. Turning the wheel turns the axle, making it easier to move loads (e.g., a steering wheel or a screwdriver).

📐Formulae

Work=Force×DistanceWork = Force \times Distance

Load×Load Arm=Effort×Effort ArmLoad \times Load\ Arm = Effort \times Effort\ Arm

W=F×sW = F \times s

💡Examples

Problem 1:

A boy applies a force of 20 N20\ N to push a box across a floor for a distance of 5 m5\ m. Calculate the work done.

Solution:

Work=Force×DistanceWork = Force \times Distance Work=20 N×5 m=100 JoulesWork = 20\ N \times 5\ m = 100\ Joules

Explanation:

Since the box moved in the direction of the force, work is calculated by multiplying the magnitude of the force (20 N20\ N) by the distance moved (5 m5\ m).

Problem 2:

In a Class I lever, a load of 50 N50\ N is placed at a distance of 2 m2\ m from the fulcrum. How much effort is needed if the effort is applied at a distance of 4 m4\ m from the fulcrum?

Solution:

Load×Load Arm=Effort×Effort ArmLoad \times Load\ Arm = Effort \times Effort\ Arm 50 N×2 m=Effort×4 m50\ N \times 2\ m = Effort \times 4\ m 100=Effort×4100 = Effort \times 4 Effort=1004=25 NEffort = \frac{100}{4} = 25\ N

Explanation:

Using the principle of levers, the product of load and its distance from the fulcrum must equal the product of effort and its distance from the fulcrum.

Problem 3:

Identify the type of simple machine used in a bottle opener and specify the positions of the Fulcrum, Load, and Effort.

Solution:

The bottle opener is a Class II Lever.

Explanation:

In a bottle opener, the Fulcrum (FF) is at the end touching the cap top, the Load (LL) is the resistance of the cap in the middle, and the Effort (EE) is applied at the handle end. Since the Load is between the Fulcrum and Effort, it is a Class II lever.