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Question

Figure shows two blocks in contact and sliding down the inclined surface of inclination 30. Coefficient of friction between the block of mass 4 kg and the inclined plane is μ1=0.3 and that between the block of mass 2 kg and the inclined plane is μ2=0.20. Assume coefficients of static and kinetic friction to be equal. Find the acceleration of 2 kg block. Take g=10 m/s2.


A
15433 m/s2
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B
53 m/s2
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C
15 m/s2
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D
5 m/s2
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Solution

The correct option is A 15433 m/s2
FBD of the system:


m1=4 kg, m2=2 kg
Since there is relative motion between the blocks and incline, kinetic friction will act.
f1=μ1N1=μ1m1gcosθ
f2=μ2N2=μ2m2gcosθ

Since, μ2<μ1, acceleration of 2 kg block down the plane will be more than the acceleration of 4 kg
Block 2 kg will always remain in contact with block 4 kg.

Applying the equation of dynamics down the inclined plane for system of (2 kg+4 kg):
Fnet ext=(2+4)anet
(m1+m2)gsinθf1f2=6anet
6gsinθμ1m1gcosθμ2m2gcosθ=6anet
6anet=6×10×12(0.3×4×1032)(0.2×2×10×32)
anet=30836
anet=15433 m/s2

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