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Work, Energy and Simple Machines - Work Done by a Constant Force

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) acting on an object 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 the object, and (2) The object must be displaced.

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The SI unit of work is the Joule (JJ). 1 J1\text{ J} is defined as the amount of work done when a force of 1 N1\text{ N} displaces an object by 1 m1\text{ m} in the direction of the force.

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Work is a scalar quantity, meaning it has magnitude but no direction.

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Work done can be positive, negative, or zero: (1) Positive if force and displacement are in the same direction, (2) Negative if they are in opposite directions (e.g., friction), and (3) Zero if there is no displacement or the force is perpendicular to the displacement (e.g., a porter carrying a load on his head and standing still).

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The general expression for work done when the force is at an angle θ\theta to the displacement is W=Fscos⁡(θ)W = F s \cos(\theta).

📐Formulae

W=F×sW = F \times s

1 Joule=1 Newton×1 metre1\text{ Joule} = 1\text{ Newton} \times 1\text{ metre}

W=m×g×hW = m \times g \times h

💡Examples

Problem 1:

A force of 15 N15\text{ N} is acting on an object. The object is displaced through 4 m4\text{ m} in the direction of the force. Calculate the work done.

Solution:

W=F×sW = F \times s W=15 N×4 mW = 15\text{ N} \times 4\text{ m} W=60 JW = 60\text{ J}

Explanation:

Since the displacement is in the direction of the force, we multiply the magnitude of the force (15 N15\text{ N}) by the displacement (4 m4\text{ m}) to get the total work done in Joules.

Problem 2:

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

Solution:

Mass of luggage, m=15 kgm = 15\text{ kg}, Displacement, s=h=1.5 ms = h = 1.5\text{ m}. Force applied equals the weight of the object: F=m×gF = m \times g 15×10150\begin{array}{r} 15 \times 10 \\ \hline 150 \end{array} So, F=150 NF = 150\text{ N}. Work done W=F×sW = F \times s W=150 N×1.5 mW = 150\text{ N} \times 1.5\text{ m} 150×1.5225\begin{array}{r} 150 \times 1.5 \\ \hline 225 \end{array} W=225 JW = 225\text{ J}

Explanation:

To lift the object, the porter must apply a force equal to the weight of the object (m×gm \times g). The work done is then calculated by multiplying this force by the height to which the luggage is raised.