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Asked by kumarabhishekuprama | 10 Aug, 2021, 17:48: PM
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Let us consider that long piece paper is moving in the direction of +x-axis as shown in fig(1). Initially small block is moving in +y-axis direction.
Once small block is entered in the paper, it has x-component and y-component of motion. Trajectory of the block is shown in fig.(2) .
 
Fig.(3) gives the free body diagram of moving block . We see that friction force f = ( μ m g ) is opposing the movement of block.
 
Where μis friction coefficient, m is mass of block and g is acceleration due to gravity
 
Retardation a experienced by block = friction force / mass
 
Retardation a = ( μ m g ) / m = ( μ g )
 
Maximum velocity v of the block , if the block does not crosses the edge of paper is determined from the following equation of motion.
 
vf2 = v2 - ( 2 a L ) 
 
where vf = 0 is the final velocity , v is initial velocity , a is retardation and L is width of paper.
 
Hence , we get , begin mathsize 14px style v space equals space square root of 2 space cross times space a space cross times L end root space equals space square root of 2 space cross times space 0.1 space cross times 9.8 space cross times fraction numerator square root of 5 over denominator 2 end fraction end root space equals space 1.48 space m divided by s end style
Answered by Thiyagarajan K | 11 Aug, 2021, 09:19: AM
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