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CBSE Class 11-science Answered

my question is on friction:
Asked by ashwati | 23 Sep, 2010, 12:17: AM
answered-by-expert Expert Answer
a/ Net force on 2 kg block = Applied - Friction = 10 - (0.2)(2)(10) = 6 N, hence a1 = 6/2 = 3 m/s2.
Now only friction force acts on block of mass 3 kg, but the friction force between 3 kg and 7 kg is, 0.3x(2 + 3)(10) = 15 N < applied i.e. 4 N (which is friction force between 2 and 3 kg block) , and therefore 3 kg and 7 kg will move together, since the there is no friction force from ground.
Hence a2 = a3 = 4/(2+3) = 4/5 m/s2.
b/If 10 N is applied at the middle block, then the friction force is 4 N + 15 N = 19 N < 10 N, hence all the three blocks will move together,
a1 = a2 = a3 = 10/(2 + 3 + 7) = 10/12 = 5/6 m/s2.
c/If 10 N is applied at the last block, then the friction force is 15 N < 10 N, hence all the three blocks will move together,
a1 = a2 = a3 = 10/(2 + 3 + 7) = 10/12 = 5/6 m/s2.
regards,
Team,
TopperLearning.
Answered by | 25 Sep, 2010, 01:26: AM

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