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Exercise · Q9

Q.Explain why the work done by a force is a scalar quantity, even though both the force and the displacement it acts through are vector quantities.

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Work is defined as W=F⃗⋅d⃗=Fdcos⁡θW = \vec F \cdot \vec d = Fd\cos\theta where F⃗\vec F is the force vector, d⃗\vec d is the displacement vector, and θ\theta is the angle between them. The operation combining the two vectors here is the dot product (scalar product), and the defining property of the dot product of any two vectors is that its result is always a single number -- a magnitude with a sign, but no direction of its own -- never another vector.

This is easiest to see from the formula itself: FF, dd and cos⁡θ\cos\theta are each ordinary numbers (the magnitudes of the two vectors, and a trigonometric ratio), so their product Fdcos⁡θFd\cos\theta is necessarily just a number too. Physically, the dot product extracts only the component of one vector that lies along the direction of the other -- here, the component of the force lying along the direction of motion -- and multiplies it by the magnitude of that motion; a "component along a direction" is inherently a scalar quantity, since the direction itself has already been used up in specifying which component is being taken, leaving only a magnitude (positive, negative or zero) behind.

This is why, even though force and displacement each need both a magnitude and a direction to be fully specified, work itself needs only a single number: a positive value (force helping the motion along), a negative value (force opposing it), or zero (force perpendicular to the motion) -- but never a "direction of work" in the way there is a direction of force or of displacement.

✓Final answer

Work is a scalar because it is the dot product F⃗⋅d⃗=Fdcos⁡θ\vec F \cdot \vec d = Fd\cos\theta of two vectors, and a dot product, by its very definition, always produces a single scalar number, never a vector.

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