Question

In the given arrangement two blocks are placed on a horizontal surface such that efficient of friction between the

Answer 1

R1-2g=0

R1=2x10=20

2a+0.2 R1-12=0

2a+0.2(20)=12

2a=12-4

a=4m/.s2

Again, 4a-μR1=0.2(20)=4

a1=1m/s2

2kg block has acceleration 4m/s2 and that of 4kg is 1m/s2

ii) R1=2g=20

ma=μ1R1=0

a=0

and 4a+0.2x2x10-12=0

4a+4=12

4a=8 m/s2

and a= 2m/s2