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Please answer this

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Asked by Prashant DIGHE 17th September 2019, 9:56 PM
Answered by Expert
Answer:
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 = begin mathsize 12px style square root of 80 end style  m/s
Since AB is smooth surface , velocity at B = vB = begin mathsize 12px style square root of 80 end style 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 = begin mathsize 12px style square root of 76 end style m/s
Since collision with vertical wall at C is elastic, block returns back with same speed begin mathsize 12px style square root of 76 end style 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' = begin mathsize 12px style square root of 72 end style m/s
Then the returning block travels the smooth surface in backward direction,  falls back
and reaches the point A with same speed begin mathsize 12px style square root of 72 end stylem/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 begin mathsize 12px style square root of 80 end style begin mathsize 12px style square root of 72 end style 6
2 begin mathsize 12px style square root of 72 end style begin mathsize 12px style square root of 64 end style 12
3 begin mathsize 12px style square root of 64 end style begin mathsize 12px style square root of 56 end style 18
4 begin mathsize 12px style square root of 56 end style begin mathsize 12px style square root of 48 end style 24
5 begin mathsize 12px style square root of 48 end style begin mathsize 12px style square root of 40 end style 30
6 begin mathsize 12px style square root of 40 end style begin mathsize 12px style square root of 32 end style 36
7 begin mathsize 12px style square root of 32 end style begin mathsize 12px style square root of 24 end style 42
8 begin mathsize 12px style square root of 24 end style begin mathsize 12px style square root of 16 end style 48
9 begin mathsize 12px style square root of 16 end style begin mathsize 12px style square root of 8 end style 54
At the beginning of 10th cycle, block starts from A with speed begin mathsize 12px style square root of 8 end style m/s,  reaches B with same speed,
reaches C after retrdation with speed begin mathsize 12px style square root of 4 end style 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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