Determine the time in which the smaller block reaches other end of bigger block in figure

Options

A. `4 s`
B. `8 s`
C. `2.19 s`
D. `2.13 s`

Answer

c

Explanation

Friction between `2kg and 8 kg` blocks in nature, so

`F = m xx 2g = 0.3xx 2xx 10 = 6N`
For `2 kg`block`10- 6= 2a_(1)`
For `8 kg` blocks `6 = 8a_(2) rArr a_(1) = 2ms^(-2), a_(2) = (3)/(4)ms^(-2)`
Acceleration of `2 kg` blocks relative to `8 kg` blocks is
`a_(rel) = a_(1) - a_(2) = (5)/(4) ms^(-2)`
Using the equation of motion `3 = (1)/(2) xx(5)/(4) r^(2) rArr r = 2.19s`