Sarahisme Posted March 26, 2005 Posted March 26, 2005 Hey just wanted to check my answer for this question damn answers in back of book only being for every 2nd question i get v = 1.40m/s well heres the question
Sarahisme Posted March 27, 2005 Author Posted March 27, 2005 ok here is what i did..... (dEmech, means delta Emech, and so on) Wext = dEmech = 0 = dKE + dPE define PE=0 where the 2kg is initially. therefore PEi = m1gh + m2gh = m1g KEi = 0 m1g = PEf + KEf = m1gh + m2gh + .5m1v1^2 + 0.5m2v2^2 = gh(m1 + m2) + .5v^2(m1+m2) m1g = (m1 + m2)(gh + .5v^2) v^2 = 2(m1g/(m1+m2)-gh) subing in values for m1, m2, g and h. h is 0.5 because of the length of the rope attaching the two blocks. (the only possible height when they can be at the same height is when h = 0.5m) so therefore v = 1.40m/s or 1m/s (because the quesiton only has one sig figure)
swansont Posted March 27, 2005 Posted March 27, 2005 Looks good to me! The reason for asking to show your work is that there are some ways to do a problem incorrectly (like making two mistakes that cancel in the math) that still give the right answer.
Sarahisme Posted March 27, 2005 Author Posted March 27, 2005 oh ok, yeah i know what ya mean well thanks for that then
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