Two masses `m_(1)` = 5 kg and `m_(2)` = 10 kg, connected by an inextensible string over a frictionless pulley are moving as shown in the figure. The coefficient of friction of horiztonal surface is 0.15. The minimum height m that should be put on top of `m_(2)` to stop the motion is

Options

A. 18.3 kg
B. 27.3 kg
C. 43.3 kg
D. 10.3 kg

Answer

C

Explanation

There is error in given option
Let mass m is placed on block `m_(2)` so that it does not moves .
Static friction force = weight of block of `m_(1)`
`f_(s ) = m_(1) g`
but ` f_(s ) = mu_(s ) (m_(2) + m) g`
`mu_(s ) (m_(2) +m) g = m_(1) g`
`0.15 (10 + m) = 5` `10 + m = (500)/( 15)`
`m= 33.33 -10`
`m= 23.3 kg`