In the other scenarios, it is easy to see what is doing negative work on the book. The only force here is gravity, which is certainly doing positive work on the book given every definition of work I've seen. In scenario A, I would argue that nothing has done negative work on the book. But what if I lower the book by one meter? That implies that:Ī.) I let the book free fall that one meter and stop looking at the question once the book hits the one meter mark.ī.) I slowly lower the book, exerting a force the entire way to keep it from free falling.Ĭ.) I let the book free fall, and quickly 'catch it' so that it comes to a halt at the one meter mark. For example, if I hold a book in my hand and raise it by one meter, then I've done work on the book obviously. Some questions have implicit and unstated conditions that aren't immediately obvious. You can choose to say, "Well, that's just convention" and be done with it, or you can acknowledge that it doesn't actually make sense when you try to figure out what's actually happening. The crux of the OP's question is answered, in my opinion, by recognizing that some of these questions and scenarios are vague and not quite technically accurate (such as what does negative work on a free-falling object solely under the influence of gravity). Raise it by 10 units of PE and gravity does -10 units of work, just as it should be given the other definitions of work. it's falling) will have W=-ΔU=-U 2-U 1=-(10-20)=-(-10)=10 units of work performed on it by gravity. A book that starts with 20 units of potential energy and ends with 10 units (i.e. In fact, the relationship between work and potential energy perfectly supports this. It has everything to do with explicitly stating which forces are involved in the question and which ones your interested in. This has nothing to do with choosing how potential is defined. In the end you have to choose which way round to specify how Potential is defined. As a consequence, the charge's potential energy also drops, in this case, becoming more negative (U = qV).īoth situations are analogous conceptually. The field is doing the work, and the field is doing the work ONTO the charge. Work done by the charge being moved is negative, because it is not doing the work. This is no different than bringing a negative charge nearer to a positive charge, or bringing a positive charge nearer to a negative charge. However, if we simply let go of the book and it free-falls to the ground, the work done ONTO the book is NEGATIVE, and consequently, it loses gravitational potential energy. the book gains gravitational potential energy. For example, lifting a book to a new height means that work is done ONTO the book, i.e. So something like this where work done is "negative" should be familiar. When someone is learning about charges, it is assume that the student has learned basic kinematics, etc. Within this site there are a number of questions about using the integral to find the work done and getting the wrong sign for the answer.Īlmost all the errors are due to using $-dr$ instead of $dr$ within the integral and/or including $\cos \pi$ when evaluating the dot product, eg link 1, link 2, etc.As a physics instructor, I often puzzle at why a student cannot make the connection with something that he/she should have learned about and understood from something earlier. With the sign of $dr$ being dictated by the limits of integration. I think you are not considering the displacement as a vector- suppose one is at a position $\mathbf\, dr$$ "we know that if the applied force is in the direction of the displacement then work done is positive.But in case of bringing 2 opposite charges from infinite to a certain distance,the work done is negative even the force and the displacement of the charge is in the same direction."
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