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Work and Energy - Scientific Concept of Work

Grade 9CBSE

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

🔑Concepts

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In physics, work is said to be done when a force (FF) acts on an object and causes a displacement (ss) in the direction of the force.

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Two conditions must be met for work to be done: (1) A force should act on an object, and (2) The object must be displaced.

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Work done is a scalar quantity; it has magnitude but no direction.

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The SI unit of work is the Joule (JJ), where 1J=1N⋅m1 J = 1 N \cdot m. One Joule is the work done when a force of 1N1 N displaces an object by 1m1 m.

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Positive Work: When the force and displacement are in the same direction (angle θ=0∘\theta = 0^\circ).

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Negative Work: When the force and displacement are in opposite directions (angle θ=180∘\theta = 180^\circ), such as work done by friction.

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Zero Work: When there is no displacement, or when the force is perpendicular to the displacement (angle θ=90∘\theta = 90^\circ).

📐Formulae

W=F×sW = F \times s

W=Fscos⁡θW = F s \cos \theta

1J=1N×1m1 J = 1 N \times 1 m

💡Examples

Problem 1:

A force of 5N5 N is acting on an object. The object is displaced through 2m2 m in the direction of the force. If the force acts on the object all through the displacement, find the work done.

Solution:

W=F×s=5N×2m=10JW = F \times s = 5 N \times 2 m = 10 J

Explanation:

Since the force and displacement are in the same direction, the work done is simply the product of force and displacement.

Problem 2:

A porter lifts a luggage of 15kg15 kg from the ground and puts it on his head 1.5m1.5 m above the ground. Calculate the work done by him on the luggage (Take g=10m/s2g = 10 m/s^2).

Solution:

Mass m=15kgm = 15 kg, displacement s=1.5ms = 1.5 m, and acceleration due to gravity g=10m/s2g = 10 m/s^2. Force F=m×g=15kg×10m/s2=150NF = m \times g = 15 kg \times 10 m/s^2 = 150 N. Work done W=F×s=150N×1.5m=225JW = F \times s = 150 N \times 1.5 m = 225 J.

Explanation:

The porter exerts a force equal to the weight of the luggage to lift it. The work done is calculated using the vertical displacement.

Problem 3:

An object is being pulled across a floor by a force of 10N10 N at an angle of 60∘60^\circ to the horizontal. If the object moves 5m5 m horizontally, calculate the work done.

Solution:

W=Fscos⁡θ=10N×5m×cos⁡(60∘)=10×5×0.5=25JW = F s \cos \theta = 10 N \times 5 m \times \cos(60^\circ) = 10 \times 5 \times 0.5 = 25 J

Explanation:

When the force is at an angle to the displacement, only the component of the force in the direction of displacement (Fcos⁡θF \cos \theta) contributes to the work done.