1) mgsin30 = F = (mu)R = (mu)mgcos30
so mu = tan30 = 1/rt3
now...
Force = ma = mgsin40 - (mu)mgcos40
a = gsin40 - (g/rt3)cos40 = 1.97ms^-2
2. A particle of mass 2kg rests in limiting equilibrium on a plane inclined at 25deg to the horizontal. The angle of inclination is decreased to 20deg and a force of magnitude 20N is applied up a line of greatest slop. Find the particle's acceleration. When the particle has been moving for 2 seconds the force is removed. Determine the distance the particle will move up the plane.
2) 2gsin25 = (mu)2gcos25
mu = tan25 = 0.466...
now friction acts in other direction...
20 - 2gsin20 - (mu)2gcos20 = Force = 2a
a = 10 - g(sin20 + tan25cos20) = 2.35ms^-2
in that time period it goes: s = ut + 1/2at^2 = 0.5 * 2.35 * 2^2 = 4.708m
velocity at end of that u + at = 4.708s.
then when it's slowing down, acceleration is g(sin20 + tan25cos20) = 7.65ms^-2, s = (v^2 - u^2)/2a = 1.45m
add them up to get 6.16m
phew. you just know there's a mistake somewhere...