Question

two blocks of masses 5kg and 3kg are placed on a smooth horizontal surface. A horizontal force of 16 N is applied on 5 kg as shown. find the ratio of normal force between the blocks and acceleration of the blocks

Answer 1

Answer:

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Step-by-step explanation:

block B is over block A, so normal force between the blocks will be weight of block B. =30N.

block B is over block A, so normal force between the blocks will be weight of block B. =30N.so maximum friction force that can be generated is f=μ.N=0.5×30=15N

block B is over block A, so normal force between the blocks will be weight of block B. =30N.so maximum friction force that can be generated is f=μ.N=0.5×30=15Nso maximum acceleration of block B is

block B is over block A, so normal force between the blocks will be weight of block B. =30N.so maximum friction force that can be generated is f=μ.N=0.5×30=15Nso maximum acceleration of block B is m

block B is over block A, so normal force between the blocks will be weight of block B. =30N.so maximum friction force that can be generated is f=μ.N=0.5×30=15Nso maximum acceleration of block B is mf

block B is over block A, so normal force between the blocks will be weight of block B. =30N.so maximum friction force that can be generated is f=μ.N=0.5×30=15Nso maximum acceleration of block B is mf

block B is over block A, so normal force between the blocks will be weight of block B. =30N.so maximum friction force that can be generated is f=μ.N=0.5×30=15Nso maximum acceleration of block B is mf =

block B is over block A, so normal force between the blocks will be weight of block B. =30N.so maximum friction force that can be generated is f=μ.N=0.5×30=15Nso maximum acceleration of block B is mf = 3

block B is over block A, so normal force between the blocks will be weight of block B. =30N.so maximum friction force that can be generated is f=μ.N=0.5×30=15Nso maximum acceleration of block B is mf = 315

block B is over block A, so normal force between the blocks will be weight of block B. =30N.so maximum friction force that can be generated is f=μ.N=0.5×30=15Nso maximum acceleration of block B is mf = 315

block B is over block A, so normal force between the blocks will be weight of block B. =30N.so maximum friction force that can be generated is f=μ.N=0.5×30=15Nso maximum acceleration of block B is mf = 315 =5m/s

block B is over block A, so normal force between the blocks will be weight of block B. =30N.so maximum friction force that can be generated is f=μ.N=0.5×30=15Nso maximum acceleration of block B is mf = 315 =5m/s 2

so minimum acceleration of both the blocks can be 5m/s

so minimum acceleration of both the blocks can be 5m/s 2

so minimum acceleration of both the blocks can be 5m/s 2 if both blocks don't want to move together.

so minimum acceleration of both the blocks can be 5m/s 2 if both blocks don't want to move together.so minimum force that required for relative motion is m.a=8×5=40N

so minimum acceleration of both the blocks can be 5m/s 2 if both blocks don't want to move together.so minimum force that required for relative motion is m.a=8×5=40Nmeans force should be less then 40N, converting to (in kg wt.)(divide by 10) we get maximum force is less then 4kg wt.

so minimum acceleration of both the blocks can be 5m/s 2 if both blocks don't want to move together.so minimum force that required for relative motion is m.a=8×5=40Nmeans force should be less then 40N, converting to (in kg wt.)(divide by 10) we get maximum force is less then 4kg wt.so from the above options we found option B as maximum but less then 4, so option B is answer.