Why is Work Calculated as Fdcosθ Even with Gravity?

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    Conceptual Work
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Work is defined as the dot product of force and displacement, represented by the equation Fdcosθ, where θ is the angle between the force and displacement vectors. This definition applies universally, including when calculating work done by gravity. Work is a scalar quantity, meaning it has magnitude but no direction. The cosine factor accounts for the angle between the force applied and the direction of movement, which is crucial for accurate calculations. Understanding this concept is essential for grasping the principles of mechanical work.
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Homework Statement



This is conceptual question...Is work a scalar or vector quantity? Is it force times distance or displacement?

And it just Fdcostheta by definition of the dot product. Why is it cos theta even when you calculate work done by gravity?

Homework Equations



Fdcos theta

The Attempt at a Solution



Just need conceptual clarification.
 
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hello,

just check http://en.wikipedia.org/wiki/Mechanical_work"

I suppose,it helps.

Dot product of two vectors is scalar :smile:
 
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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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