Recent content by azn_oohwee

oops sorry.. I meant to write.. ∑(Fb)x = -T - fk - mgsin(@) = ma ∑(Fb)x = -T - ukN - mgsin(@) = ma N = mgcos@ so ∑(Fb)x = -T - ukmgcos@ = ma -m(a)g - uk(m(b)gcos@ - m(b)gsin@ = m(b)a + m(c)a a = - { g(m(c) - uk(m(b)cos@) - m(b)sin@ } / {m(b) + m(c) } That should be more...

I am having a problem finding the last part of this problem, and I'm not really sure if I did the first part right. The 1.0 kg physics book in Figure P8.38 is connected by a string to a 600 g coffee cup. The book is given a push up the slope and released with a speed of 3.0 m/s. The...