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I would like to know why two forces (of different magnitudes but same direction) that we'd use sequentially on an object to move it by the same amount twice, will produce two different values for the work quantity.

Why is there a mismatch between the maths and the observed physical change?

What exactly is the work of a force? (If possible, please avoid circular definitions like "work is the energy produced by a force" and "energy is the ability to do work".)

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I would like to know why two forces (of different magnitudes but same direction) that we'd use sequentially on an object to move it by the same amount twice, will produce two different values for the work quantity. [...] Why is there a mismatch between the maths and the observed physical change?

In $W=Fd$, the distance $d$ should not be interpreted as the motion caused by the force. In general, force doesn't cause motion, it causes acceleration. The physical change that is described by work is the change in the object's energy, not the change in its position. This explains the apparent mismatch.

In your example, the weaker force could be a zero force. The observed physical change in the object is zero, because the object's energy doesn't change. The object continues in the same state of motion.

If possible, please avoid circular definitions like "work is the energy produced by a force" and "energy is the ability to do work"

See the answers to this question: Intuitively Understanding Work and Energy

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When the force acts on a body, it displaces the body through a certain distance in the direction of force or against it.

$W = \vec{F} \cdot \vec{s} = |F| |s| \cos \alpha$

Where, $W$ is the work done by a force $\vec{F}$ causing a displacement $\vec{s}$ and $\alpha$ is the angle between the force and the displacement vectors.

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protected by Qmechanic Mar 7 '14 at 7:37

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