How to Solve Work Problems: Understanding the Concept and Calculations

[XenoFazz]

Hey,

Rather confused about a concept with "Work". Say you're given an inclined plane problem where you're told to find the work energy required to move an object to the top of the plane (this is a TYPE of work problem, I should think).

The answer (anyhow), in value is equal to the gravitational potential energy (mgh). Now... I ask. Why is that? (In other words, why isn't it the applied force (minus Fg parallel or Fg sin theta) - could you not consider finding work using the applied force PLUS using the length of the incline (hypotenuse) as delta d instead?)

This problem has shown up in the past, the absolute BEST you could do is just clarify "conceptually". Or just help. Thanks!

 

[cepheid]

Calculating mgh gives you the minimum amount of work needed to raise the object to that height. You could very well do more work to get it there, if you were so inclined (ohhh...terrible pun). For instance, if you wanted to accelerate the object, or if there were friction on the ramp, then you would end up doing more work to achieve the same result. However, if you do the calculation for a frictionless incline in the scenario in which there is no net force on the object (ie the applied force just counteracts gravity and the object moves up the ramp at a constant velocity) then you should get the same answer as mgh. Try it!

That's the clearest explanation I could think of. For more detail...does this mean anything to you: if the net force on the object is zero, then the work you do on the object is equal to the negative of the work done by the graviational force on that object throughout its displacement. Since the graviational force is a conservative vector field, the work it does on an object moving from point a to point b is independent of the path taken by the object between those two points. If this does not mean anything yet, don't worry about it.

 

[wais]

Hi there,

I can definitely understand your confusion with work problems, especially when it comes to inclined planes. To better understand the concept of work, it is important to first define what work actually is. In physics, work is defined as the product of force and displacement in the direction of the force. This means that in order to do work on an object, you need to apply a force and cause it to move a certain distance in the direction of the force.

Now, let's apply this concept to the inclined plane problem you mentioned. You are correct in saying that finding the work energy required to move an object to the top of the plane is a type of work problem. In this case, the work being done is against the force of gravity, which is pulling the object down the incline. The work done against gravity is equal to the change in gravitational potential energy, which is given by the formula mgh (mass x gravity x height). This is because as the object moves up the incline, it gains potential energy due to its increased height.

So why isn't the work equal to the applied force minus the force of gravity parallel to the incline? This is because the force of gravity parallel to the incline does not actually do any work. It only affects the normal force, which is perpendicular to the incline and does not contribute to the work being done. Therefore, when calculating work, we only consider the force that is actually causing the object to move, which is the force of gravity pulling it down the incline.

I hope this helps to clarify the concept of work and how it applies to inclined plane problems. Remember, when solving work problems, always think about the direction of the force and the displacement of the object in that direction. Good luck with your studies!