Recent content by murcy

Thanks a bunch Jay~

Thanks for the reply Jay; I understand what you mean now, this is what i have so far i hope is right; Wfriction= (1/2)(m1+m2+m3)v^2+(m1-m3)gΔy Wfriction=(1/2)(0.8kg+1.1kg+1.2kg)v^2+(0.8kg-1.2kg)(9.8m/s^2)(0.6m) Wfriction=1.55kg*v^2-2.352joules earlier i found; Wfriction=fdcosθ...

I'm sorry but I am still stuck in this problem :/

Hi and thanks :), sorry about the upper case M.(changed).ok so the work on m2 would be; w= (1/2)m2v^2 frcosθ= (1/2)m2v^2 N*μ(-1)=(1/2)m2v^2 -mg*μ=(1/2)m2v^2 -(1.1kg)(9.8m/s^2)*(0.345)=(1/2)(1.1kg)v^2 solve for V; -3.719N=0.55kg v^2 v^2=-3.719N/0.55kg v^2=-6.762m^2/s^2 v=sqrt(-6.762...

Problem and Data three masses conected m1= 800g, m2 =1100g m3=1200g. m2 rest on a table of coefficient μ=0.345. m1 and m3 hang vertically over frictionless/massless pulleys. The system is released from rest. Whats the velocity of m3 after falling 60.0cm P.S "i know this system can be...