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

Options

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

Answer

D

Explanation

Let `mu_s` and `mu_k` be the coefficients of static and kinetic friction between the box and the plank respectively.
when the angle of inclination ` theta ` reaches `30^@` , the black just slides ,
` therfore mu_s `a is the acceleration
produced in the black , then

ma `= mg sin theta- f_k`
(where `f_k` is force of kinetic friction )
`=mg sin theta - mu _k N " " ( "as" f_k = mu_k N)`
` = mg sin theta - mu_k mg cos theta ( " as" N= mg cos theta)`
` a= g ( sin theta - mu_k cos theta ) `
As ` g= 10 ms^(-2) and theta = 30^@`
` therefore a= ( 10 ms^(-2)) ( sin 30^@ - mu_k cos 30^@ )`
if s is the distance travelled by the black in time t,then
` a= ( 2s )/( t^2) ` ( as u=0)
but s= 4.0 m and t = 4.0 s ( given )
` therefore a=( 2 ( 4.0 m ) )/(4.0 s)^(2 ) = 1/2 ms^(-2)`
substiting this value of a in eqn. (i ) we get
`1/2 ms^(-2) = (10 ms^(-2)) ((1)/(2) -mu_k ( sqrt(3))/(2)), mu_k = ( 0.9 )/( sqrt(3)) = 0.5 `