Question

Block A of weight 100 kg rests on a block B and is tied with horizontal string to the wall at C. Block B is of 200 kg. The coefficient of friction between A and B is 0.25 and that between B and surface is $\frac{1}{3}$. The horizontal force F necessary to move the block B should be (g = 10 m/s²)
(a) 1050 N
(b) 1450 N
(c) 1050 N
(d) 1250 N

Answer 1

The answer will be option (d) 1250 N

Here is the explanation...

Friction between A & B is 0.25 and that between B and surface is 1/3...

so from the formula...

P = F(AB) + F(BS)  

= μ(AB)m(Ag) + μ(BS) [m(A) + m(B)]g  

= 0.25 × 100 × 10 + 1/3 [100 + 200] × 10 = 1250 N

So the horizontal force F necessary to move the block B should be 1250 i.e. option (d).

Answer 2

Answer:1250N

Explanation: