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Two blocks of mass m1 and m2 are connected by a weightless rod on a plane having inclination of 37. The coefficients of dynamic friction of m1 and m2 with the inclined plane are μ=0.25. Then the common acceleration of the two blocks and the tension in the rod are

  1. 4 m/s2,T=0
  2. 2 m/s2,T=5 N
  3. 10 m/s2,T=10 N
  4. 15 m/s2,T=9 N


Solution

The correct option is A 4 m/s2,T=0
From FBD of m2


m2gsinαμm2gcosαT=m2a   ... (1)
From FBD of m1
m1gsinα+Tμm1gcosα=m1a   ... (2)
Solving (1) and (2) for the value of T,
m1m2gsinαμm1m2gcosαm1T=m1m2gsinα+m2Tμm1m2gcosα
T(m1+m2)=0
T=0           ... (3)
Substituting the value of (3) in either (1) or (2) we get
a=gsinαμgcosα
a=10 sin37(0.25×10cos37)
a=4 m/s2

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