+2 votes
in JEE by kratos

Two blocks of masses 20 kg and 10 kg are kept or a rough horizontal floor. The coefficient of friction between both blocks and floor is μ= 0.2. The surface of contact of both blocks are smooth. Horizontal forces of magnitude 20 N and 60 N are applied on both the blocks as shown in figure. Match the statement in column-I with the statements in column-II.

| Column-I | Column-II |
| (A) Frictional force acting on block of mass 10 kg | (p) has magnitude 20 N |
| (B) Frictional force acting on block of mass 20 kg | (q) has magnitude 40 N |
| (C) Normal reaction exerted by 20 kg block on 10 kg block | (r) is zero |
| (D) Net force on system consisting of 10 kg block and 20 kg block | (*) is towards right (in horizontal direction). |

1 Answer

+4 votes
by kratos
 
Best answer

*Correct option: (A) p, (B) p, (C) q, (D) r**

Explanation:

The minimum horizontal force required to push the two block system towards left

= 0.2 × 20 × 10 + 0.2 × 10 × 10 = 60.

Hence the two block system is at rest. The FBD of both of blocks is as shown. The friction force f and normal reaction N for each block is as shown.

Hence magnitude of friction force on both blocks is 20 N and is directed to right for both blocks. Normal reaction exerted by 20 kg block on 10 kg block has magnitude 40 N and is directed towards right. Net force on system of both blocks is zero.

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