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Question

A system of 3 blocks of masses of 10 kg,20 kg,30 kg respectively, having flat surfaces of contact with coefficients of friction given in the figure are at rest initially. If a constant horizontal force of F=100 N is acting on 10 kg block as shown in figure, then the magnitude of acceleration of the 20 kg block with respect to the ground is (Take g=10 m/s2)

A
1 m/s2
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B
2 m/s2
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C
3 m/s2
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D
5 m/s2
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Solution

The correct option is A 1 m/s2Let as assume the blocks are moving together as one system with common acceleration ac. FBD of 10 kg block: From figure (i), we get R1=10g=100N 100−f1=10ac−(1) FBD of 20 kg+30 kg: From figure (ii), we get R′1=R1+50g and f1=50ac−(2) Solving (1) & (2), 100−50ac=10ac ac=10060=1.66 m/s2 ⇒f1=50×1.66=83.33 N But, f1max=μ1R1=0.5×100=50 N Since required f1>(f1)max,10 kg block slips on the system of the other two blocks and the kinetic friction is only 50 N. Therefore, accelaration of 10 kg block a1=F−f1m=100−5010=5 m/s2. To check if there is slip between 20 kg and 30 kg blocks, draw the FBDs From (iii), R2=R1+20g=300 N and f1−f2=50−f2=20a−−(3) From (iv), R3=300+R2=600 N and f2=30a−−(4) Solving (3) & (4), we get 50=50a⇒a=1 m/s2 ⇒f2=30a=30 N f2max=μ2R2=0.25×300=75 N Since f2<(f2)max, there is no slipping between 20 kg & 30 kg blocks ∴ They both move together with an acceleration a=1 m/s2 w.r.t ground.

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