In the Atwood machine shown below, m 1 = 2.00 kg and m 2 = 7.90 kg. The masses o
ID: 2239440 • Letter: I
Question
In the Atwood machine shown below, m1 = 2.00 kg and m2 = 7.90 kg. The masses of the pulley and string are negligible by comparison. The pulley turns without friction and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at vi = 2.30 m/s downward.
magnitude 2 m/s direction 3---Select---upwardvelocity is zerodownwardExplanation / Answer
(a)Let the heavier mass displacement upward by h meter,By the law of energy conservation:- For heavier mass:- =>PE(initial) + KE(initial) = PE(final) =>mgh(initial) + 1/2mvi^2 = mgh(final) =>vi^2 = 2g[h(final) - h(initial)] =>delta h = (2.3)^2/(2 x 9.8) = 0.26m The displacement will be same for lighter mass in downward direction. (b) By v^2 = u^2 - 2as =>0 = (2.3)^2 - 2 x a x 0.26 =>a = 10.1 m/s^2 By v = u - at =>0 = 2.3 - 10.1 x t =>t = 0.31 sec =>after 0.31 sec the masses will stop momentarily and than will start to move under 'g', Thus:- =>m2g - T = m2a ----------(i) =>T - m1g = m1a ---------(ii) By (i) + (ii) :- =>a = [(m2-m1)g/(m1+m2)] = [(7.9 - 2) x 9.8/(7.9 + 2)] = 5.84 m/s^2 Let the velocity of m1 after (1.80 - 0.31) = 1.49 sec is v m/s in upward direction, =>By v = u + at =>v = 0 + 5.67 x 1.49 = 8.45 m/s(upward)