Recent content by Zobrox

will post shortly... just editing So first we know that velocity is the rate that distance changes, with respect to time: v = \frac{dx}{dt} if we rearrange the above, we get: (1) vdt = dx Now... we can represent the change in distance as (2) dx = x-x_{0}, where x_{0} is the initial distance...

Hmm, so your saying I shouldn't deduct the work due to weight with respect to block B? I thought that since the weight is opposing block B's motion this would be a negative work? I added the work due to weight from block B just to see, and it still doesn't yield the correct answer.

Thanks for your response haru, I feel it does answer my question... however I still am finding it hard to grasp the sign convention for everything... I have chosen up as positive for block B and down as positive for block A. This leads to a positive displacement for both block A and B right? so...

Okay, I'll write up my working now. I could approach it using kinematics, however it bugs me that I don't have a firm understanding on work energy, so would like to solve it this way :) Hope you can spot my errors when I update, thanks

Hi guys, really confusing myself over grasping this principle... I think my main problem is understanding datums and how choosing the direction of a positive axis effects the result. I understand that the work energy principle states: Wnet = ½m(vf2+vi2) I have previously been taught that Wnet...