Block A weighing 100 kg rests on a block B and is tied with a horizontal string to the wall at C. Block B weighs 200 kg. The coefficient of friction between A and B is 0.25 and between B and the surface is 1/3. The horizontal force P necessary to move the block B should be `(g=10m//s^(2))`

Options

A. 1050 N
B. 1450 N
C. 1050 N
D. 1250 N

Answer

D

Explanation

`F_(1)` = Froce of friction between B and A `= mu_(1) m_(1) g`
`=0.25 xx 100 xx g = 25` g netwon.
`= mu_(2) m_(2) g = mu_(2)`(mass of A and B)g
` =(1)/(3)(100+200)g =(300)/(3)g = 100g` newton
`= mu_(2) m_(2) g = mu_(2)`(mass of A and B)g
` =(1)/(3)(100+200)g =(300)/(3)g = 100g` newton `therefore F = F_(1) + F_(2)`
` = 25g + 100g = 25g = 125 xx 10N`
`therefore F = 1250N`