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Two blocks connected by a massless string slide on an inclined plane having an angle of inclination of 37. The masses of the two blocks are M1=4 kg and M2=2 kg respectively and the coefficients of friction of M1 and M2 with the inclined plane are 0.75 and 0.25 respectively. Assuming the string to be taut, find (i) the common acceleration of two masses and (ii) the tension in the string. (sin 37=0.6, cos 37=0.8)  [Take g=9.8 ms2]


A
a=2.3 ms2, T=8.45 N
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B
a=1.3 ms2, T=5.24 N
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C
a=1.35 ms2, T=11.0 N
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D
a=9.4 ms2, T=12.5 N
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Solution

The correct option is B a=1.3 ms2, T=5.24 N

Let a be the common acceleration of the two masses M1 and M2 and T be tension of the string. 
Equation of motion for M1 moving down M1g sin θ+Tf1=M1a

since f1=μ1M1 g cos θ

M1 g sin θμ1M1 g cos θ+T=M1a(1)

Equation of motion for M2 moving down

M2 g sin θf2T=M2a

since f2=μ2M2 g cos θ

M2 g sin θμ2M2 g cos θT=M2a(2)

solving (1) and (2)

a=(M1+M2)g sin θg cos θ(μ1M1+μ2M2)(M1+M2)

=(4+2)×9.8×0.69.8×0.8(0.75×4+0.25×2)6

=1.3 ms2

T=M2 g sin θμ2M2 g cos θM2a

=2×9.8×0.60.25×2×9.8×0.82×1.3

=5.24 newton

Physics

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