A plank with a box on it at one end is gradully raised about the other end. As the angle of inclination with the horizntal reaches `30^(@)` , the box starts to slip and slide `4.0m` down the plank in `4.0s` . The coefficients of static and knitic friction between the box and the plank will be, respectively.

Options

A. `0.4` and `0.3`
B. `0.6`and `0.6`
C. `0.6` and `0.5`
D. `0.5` and `0.6`

Answer

c

Explanation

When box just slips then Coefficient of static friction
`mu_(s) = tan30^(@) = (1)/sqrt3 ~~0.6`
When box slides down the plane, coefficient of kinetic friction will be involved Let a be the acceleration of the box sliding down the plane then using
`S = ut + (1)/(2)at^(2)` we have
`4 = 0 + (1)/(2) xxa xx 4^(2)` or `a=(2)/(4) = 0.5 ms^(-2)`
Here, `a =g sin theta -mu_(k) g cos theta`
`0.5 =9.8xx sin30^(@) -mu_(k) xx 9.8 cos 30^(@)`
`0.5 =9.8 xx(1)/(2) -mu_(k) xx 9.8 xx sqrt3/2`
`0.5 =4.9 -mu_(k) xx4.9sqrt3`
`mu_(k) =(4.9 -0.5)/(4.9sqrt3) =(4.4)/(4.9 xx1.73) ~~0.5` .