when the block reaches A from top of curved surface, all its potential energy is converted to kinetic energy
Hence its speed vA at A :- mgh = (1/2) m vA2 or vA = (2gh)1/2 = (2×10×4)1/2 = m/s
Since AB is smooth surface , velocity at B = vB = m/s
In the path BC, due to friction, moving block is subjected to retardation μg = 0.1×10 = 1 m/s2
Hence when the block reahes C, its speed is calculated as,
vC2 = vB2 - (2 a S ) = 80 - (2×1×2 ) = 76 or vC = m/s
Since collision with vertical wall at C is elastic, block returns back with same speed m/s
and again it is subjected to retardation 1 m/s2.
Hence when the block reaches back at B, its speed vB' is calculated as ( vB')2 = 76 - (2×1×2) , or vB' = m/s
Then the returning block travels the smooth surface in backward direction, falls back
and reaches the point A with same speed m/s
Let us call the movement A to B, B to C, then backward direction C to B, B to A as one cycle.
At end of every cycle, the difference between square of initial speed and square of final speed is 8 m2/s2 .
Speed and horizontal distance covered at end of every cycle is listed in the following table
At the beginning of 10th cycle, block starts from A with speed m/s, reaches B with same speed,
reaches C after retrdation with speed m/s , returns back and stops at B.
Horizontal distance covered in 10th cycle is 5 m.
Hence, total horizontal distance covered by the block = 54+5 = 59 m