Okay so I am just basically double checking my work. Problem. A 2kg block is held from a massles rope of height .3m , and is aloud to swing as a simple pendulum. Once it reaches the bottom of the the swing it is released. coming to a rest of displacement d. the kinectic friction uk= .6 using conservation of energy what is speed v when it reaches the table top? b) find D What I did. I calculated work from gravity Wg= F D => (2)(9.8) (.3) = 5.88 J I used an equation w= 1/2 mv^2 5.88= 1/2 (2)(v)^2 v^2=5.88 v= 2.42 m/s Solving for Part B F=ma 5.88=(2)(a) a= -2.94 I then used Uk and fn to get friction force ... .6(9.8)(2) = 11.76 n I used my chapter 2 equations to then find D V^2= vo^2 + 2A (xf-xi) 0= 2.42^2 + 2(A) (x) x= .99 My doubts... I am confused on how to find acceleration. I know that F=ma but wouldnt 5.88 be a force because its finding how much work is being done, assuming there is no air drag it shows how much force is being excerted at the very bottom.??