Two blocks are connected by a light cord. One block, of mass 4kg, sits on a hori
ID: 1904219 • Letter: T
Question
Two blocks are connected by a light cord. One block, of mass 4kg, sits on a horizontal table with static and kinetic coefficients of friction of 0.6 and 0.4, respectively, whereas the other block, of 2 kg mass, hangs over a frictionless light pulley. The blocks are released from rest. (a) Draw a carefully labeled free body diagram for both blocks and, by using Newton's laws, show that they do not move. (b) If the two blocks are exchanged, so that the 4 kg is now the hanging block and the 2 kg sits on the table, find their acceleration now. (c) In words, state why the tension in the cord is equal to the weight of the hanging block in part(a) but not in part(b). (d) What is the minimum mass that one needs to add to the 2 kg block in part(b) for it to remain at rest when released?Explanation / Answer
(a) Draw a carefully labeled free body diagram for both blocks and, by using Newton's laws, show that they do not move.
m g = 2 * 9.8 = 19.6 N
fs = M g = 0.6 * 4 * 9.8 = 23.52 N
They do not move because 19.6N is less than 23.52N.
(b) If the two blocks are exchanged, so that the 4 kg is now the hanging block and the 2 kg sits on the table, find their acceleration now.
a = F/(m+M) = (Mg - m g)/(2+4) = (4*9.8 - 0.4*2*9.8)/(2+4) = 5.23 m/s2
(c) In words, state why the tension in the cord is equal to the weight of the hanging block in part(a) but not in part(b).
at part a the hanged block is at rest and therefore F is zero. but at part b the hanged block is moving and teherefore F is not zero.
(d) What is the minimum mass that one needs to add to the 2 kg block in part(b) for it to remain at rest when released?
F = 0
Mg - (m+m') g = 0
M = (m+m')
4 = 0.6 * (2+m')
m' = 4.67 Kg