A 100kg block slides from rest down a 30 degree slope AB=500m and then down a 20 degree slope BC =500m. the

coefficient of friction along both AB and BC=0.05. calculate the speed of the block at B and C .
the coefficient friction along the horizontal plane DC is 0.15 . what is the length ,s ,of DC if the block comes to rest at D .
can anyone please help me in solving this question ?

Quite simple.
We know the force acting on body on a plane due to gravity is mg.sin theta where theta is then angle of inclination.
So force acting on body due to gravity = 100 * 9.8 * sin(30) = 490N

Reactive force acting perpendicular to plane = mg. cos theta = 100 * 9.8 * cos 30 = 848.7N
We can compute force due to friction as coefficient of friction * force perpendicular to plane
=> 848.7 * 0.05 = 42.44N
Resultant force is Force due to gravity - Force of friction = 490 - 42.4 = 447.6N

Acceleration on body, a = F/m = 447.6 / 100 = 4.48 m/s^2

We now use equation of motion v^2 = u^2 + 2as
s = 500 u = 0 a = 4.48
so we get v^2 = 0 + 2(4.48)(500)
=> v = sqrt(4480) = 66.9 m/s which is velocity at B

Same goes for BC, this time theta is 20
So force acting on body due to gravity = 100 * 9.8 * sin(20) = 335.2N

Reactive force acting perpendicular to plane = mg. cos theta = 100 * 9.8 * cos 20 = 920.9N
We can compute force due to friction as coefficient of friction * force perpendicular to plane
=> 920.9 * 0.05 = 46N
Resultant force is Force due to gravity - Force of friction = 335.2 - 46 = 289.2N

Acceleration on body, a = F/m = 289.2 / 100 = 2.9 m/s^2

We now use equation of motion v^2 = u^2 + 2as
s = 500 u = 66.9 a = 2.9
so we get v^2 = 66.9^2 + 2(2.9)(500)
=> v = sqrt(7380) = 85.9 m/s which is speed at C

We continue from there to find distance CD.
Since D is a horizontal plane, there will be no force due to gravity. The only force acting on it will be frictional force. So we calculate acceleration (deceleration) due to friction.

Reactive force acting perpendicular to plane = mg = 100 * 9.8 = 980N