A rough inclined plane (B C E ) of height ( ( 25 6 ) m ) is kept on a rectangular wooden block (A B C D ) of height (10 ~m ), as shown in the figure. A small block is allowed to slide down from the top (E ) of the inclined plane. The coefficient of kinetic friction between the block and the inclined plane is ( 1 8 ) and the angle of inclination of the inclined plane is ( ⁻¹(0.6) ). If the small block finally reaches

Step-by-step solution

According to question, a small block is slide down from top (E ) of inclined plane as shown in figure, Force equation of a block, ( m g -f=m a ) ( ) friction force applied one block, (f= _k R ) or (f= _k R(m g ) ) (From figure) where, ( _k= ) coefficient of kinetic friction from Eq. (i), we get ( aligned & m g - _k m g = ma & a=10 -1 / 8 10 (ii) ( _k= 1 8 , given ) aligned ) ( ) Given, angle of the inclined plane, ( = ⁻¹(0.6) ) or ( =0.6 ) ( array rlrl & = 1- ^2 = 1-(0.6)^2 & =0.8 array ) Form Eq. (ii), (a=10(0.6)-