Recent content by daniellionyang

6/13

9/12 first try woot! My brother got 4/12

Thanks! You mentioned that we ignore it "in this case." Are there any cases where we do not ignore this friction?

So, in rolling without slipping, even if there is friction, we ignore it? (this means it doesn't affect the net force right?)

This is from the 2014 F=ma exam: The maximum torque output from the engine of a new experimental car of mass m is τ. The maximum rotational speed of the engine is ω. The engine is designed to provide a constant power output P. The engine is connected to the wheels via a perfect transmission that...

If you disregard the friction of the water, isn't the water stationary? Also, if the water freezes, the bob has to expand right? since the period of a pendulum is 2pisqrt(l/g), there would be no difference. However, if the bob expands, that would slightly change l.

Ok! Thanks! All of you helped a lot to allow me to understand a question that had been driving me crazy for a while!

Ah... I see... Well, thank you all for clearly up my conceptuals regarding work!

Oh... I see! Thanks! Now, can you answer the question of why F has to always equal 10? Can't you simply apply any force you want while lifting the object? (the math is in post 18.

In actuality, I did read the other post. I guess I need to be a little for clear with my question.Although Ken G said that -deltaU is for a single conservative force, consider a push to the right 1 meter. of an object resting on the table. There is no change in potential energy, but there is...

Oh wait... Yeah! I forgot that potential is for a specific force. Can you explain to me specifically in what cases -deltaU and deltaKE gives you the work of a specific force and when it gives you total work? Also, why must F be 10. Can't you just apply any force to the object you are lifting...

Firstly, since the work is 10, W=F(gravity)*x+F(force)*x=>10=-10+F, so isn't F=20. If this is so, isn't it possible to choose the magnitude of the force you apply to the object? Why does the force HAVE to be g+10? Sorry if I am being rally naive about this topic.

Yes. However, what if the final velocity is not 0. Then the extra KE would just cause a bigger final velocity which won't influence the PE. Thus, even if you have extra KE, the work still should be 10.

Oh... since Ui would be 0 and Uf =Ug+UF=0. Ok! So for W=KE, if we knew the individual deltaKE for the force, it would be 10. I have one last question. W=Fx also. Thus, assuming F is constant, W done by the force would be W=F*1=F. Thus, work depends on F. However, W=-deltaU=10 gives us that it is...

Wait, so W=deltaKE gives total work and work=-deltaU gives work by a specific force?