The friction coefficient between the table and the block shown in figure is 0.2. Find the tensions in the two strings.

Options

A. `17:24`
B. `34:12`
C. `2:3`
D. `3:2`

Answer

A

Explanation

Net pulling force `F_(1) = 15 kg - 5 kg`
`F_(1) = 10 g = 100N`
Net stopping force `F_(2) = 0.2 xx 5 xx 10 = 10 N`
Since `F_(1) gt F_(2)`, therefore the system will move (anticlockwire) with an acceleration,
`a = (F_(1) - F_(2))/(10 + 5 + 5) = (90)/(25)`
`= 3.6 m//s^(2)`
`15 kg` block
`15 xx 10 -T_(2) = 15 xx a = 15 xx 3.6`
`T_(2) = 96 N`
`5 kg` block
`T_(2) - 5 xx 10 = 5 xx a = 5 xx 3.6`
`:. T_(1) = 68 N`
`:. (T_(1))/(T_(2)) = (68)/(96) = (17)/(24)`