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Question

Two blocks m1=4 kg and m2=2 kg are connected by a weightess rod on a plane having inclination of 37. The coefficient of dynamic friction for blocks m1 and m2 with the inclined plane is μ=0.2. Assume
μk=μs=0.2 and g=10 ms2. Then the common acceleration of the two blocks and the tension in the rod will be:



A
a=4.4 m/s2,T=0
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B
a=2 m/s2,T=5 N
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C
a=10 m/s2,T=10 N
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D
a=15 m/s2,T=9 N
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Solution

The correct option is B a=4.4 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 Tension T
T(m1+m2)=0
T=0(3)
Substituting the value of (3) in either (1) on (2) we get
a=gsinαμgcosa
=gsin37μgcos37
=4.4 m/s2
 

Physics

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