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Asked by Prashant DIGHE 17th September 2019, 9:56 PM
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

 Cycle No. Initial speed (m/s) Final speed (m/s Cumulative Distance Covered ( m ) 1 6 2 12 3 18 4 24 5 30 6 36 7 42 8 48 9 54
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

Answered by Expert 18th September 2019, 11:11 AM
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Tags: distance