A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.20. Find the acceleration of two blocks if a horizontal force of 12 N is applied to (a) the upper block , (b) the lower block. Take g=10 m/s².Concept of Physics - 1 , HC VERMA , Chapter 6:Friction "
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Solution :
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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
****************************************************
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
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21
Answer:
hey mate this answer is for lower block
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