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Question

Two blocks of mass m1=4 kg and m2=2 kg, connected by a weightless rod on a plane having inclination of 37 is shown in figure. If the coefficient of dynamic friction is μ=0.25, then the common acceleration of the two blocks and the tension (T) in the rod are



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

The correct option is B 4 m/s2, T=0 N
Both the blocks are connected with the rod so both the blocks will moves with common acceleration. Simultaneously considering blocks and rod as a system.
F.B.D. of system,


From the F.B.D,
N2=m2gcosθ  and  N1=m1gcosθ
f2=μN2  and  f1=μN1
f2=μm2gcosθ           
and f1=μm1gcosθ
Total frictional force on the system,
f=f1+f2=μgcosθ(m1+m2)

Assuming a is the common acceleration along the surface.

(m1+m2)gsinθ(m1+m2)μgcosθ=(m1+m2)a
10×sin370.25×10×cos37=a
10×350.25×10×45=a
a=4 m/s2

Now for tension take m2 block as a system F.B.D. m2 Block

f=μm2gcosθ
Therefore, 
m2gsinθfT=m2a
m2gsinθμm2gcosθT=m2a
2×10×350.25×2×10×45T=2×4
124T=8
T=0 N

Physics

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