Someone derive this

[HistoRocks]

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W = Fs cos theta. Spefically, the cos thetha part. The F s makes sense... you apply a force, and the object moves some distance. The product of these then equals the work done. But the cos theta? For a flat surface it makes sense... if the object moves linearly, then only a 90 degree, 180 degree, or 0 degree angle will result - relative to the x-axis. But for the inclined plane, the work = Fs cos theta, when in other cases, you have Fs sin theta down the plane (parellel component). I guess you see my confusion. Feel free to use an integral or ODE if necessary.

 

[DrBowtie]

This is the inherent problem with non-calc based physics. They try to solve the equations for general situations instead of specific ones. The only thing you should remember is that W=F(perp)*s. You should use sine and cosine accordingly.

 

[HistoRocks]

This is the inherent problem with non-calc based physics. They try to solve the equations for general situations instead of specific ones. The only thing you should remember is that W=F(perp)*s. You should use sine and cosine accordingly.

Yeah, non-calc based physics is really out there when it comes to generalizations. I'm only taking it for the sake of GPA (can't be bothered competing with engineering pplz for curved grades). Theres a lot of stuff the book attempts to explain "intuitively" (it actually states this explicitly); a.k.a moment of inertia, angular velocity, etc. Most recently, I couldn't find a proof of Pascal's Principle. Anyway... no time to relearn the Calc I've forgotten until after May 5. What a pain. So have to rely on "intuition" until then. 👎

 

[crazy_cavalier]

W = Fs cos theta. Spefically, the cos thetha part. The F s makes sense... you apply a force, and the object moves some distance. The product of these then equals the work done. But the cos theta? For a flat surface it makes sense... if the object moves linearly, then only a 90 degree, 180 degree, or 0 degree angle will result - relative to the x-axis. But for the inclined plane, the work = Fs cos theta, when in other cases, you have Fs sin theta down the plane (parellel component). I guess you see my confusion. Feel free to use an integral or ODE if necessary.

Conceptually, there are two axes that you are interested in: one is the axis that describes the actual movement of the object being pushed (or pulled.) the other is the axis in which the force is applied.

Ultimately, do whatever manipulation is necessary to deconstruct the applied force into vector components that match the axis for which the object moves. On an inclined plane, this axis will be along the incline: whatever force applied can be expressed as the vector sum of a force in the direction of the incline (i.e. parallel) and perpendicular (or "normal") to the incline. You take the force acting along this vector and *that* magnitude will be used in your equation: W = F*d.

There is NO need for calculus or ODE's!! Okay, sure, work is the area under the curve that describes F vs distance when you plot it out, and therefore W = integral(F, dx.) BUT you can just think of it conceptually. The Force is only along the direction of movement, hence the usual cos theta idea (i.e. when on a falt surface, the cos theta refers to the angle the applied force makes to the flat surface. Similarly on an incline, the cos theta refers to an angle that is between the force applied AND THE INCLINE.)