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Suppose force is applied on an object at an angle theta and the block moves for some distance along x axis

I understand that we take x component of force which is Force cos theta and multiply it with displacement vector. Why are we ignoring the y component here? Is it because the block moves along x direction?

What if force is applied at an angle theta to the ground and the block moves at an angle gamma to the ground. What will be the work done then?

Should it be F.d. cos (theta + gamma) ?

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You are correct in answering most of your own questions. The Work done by the force on the object is equal to the component of the force that is tangent to the trajectory, i.e. along the path of motion. This turns out to be equal to dot(F, dx) where both inputs are vectors. This definition holds even for curved paths but you need to integrate along the path. This can be derived from Newton's laws, along with the definition of Kinetic Energy, by taking the dot product of both sides with v (velocity) and doing some change of variables. And yes, if both the force and the displacement make angles w.r.t the horizontal the correct angle to use is the sum or difference depending on which gives you the correct angle. I think in standard format you should have the difference between the two angles.

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  • $\begingroup$ I guess that depends on how the force is applied. Am I right? If its a push, its gamma + theta and if its a pull, its gamma - thetha $\endgroup$ – MrRobot9 May 16 at 0:38
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    $\begingroup$ In that respect, a figure would help. As long as the angle is really the smaller angle between the two arrows with the tails at the same point. $\endgroup$ – ggcg May 16 at 1:01
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It is just the matter of definition that work is defined as the scalar product of force and displacement.

In the second part of your question the work done would be F.dscos( theta - gamma).

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