A small box resting on one edge of the table | vedclass

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(A) For the block,the final velocity $v = 0$,time $t = 2 \, s$,and displacement $s = 1 \, m$.Using the equation of motion $v = u + at$,we get:$0 = u +

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must be less than $0.05$

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(A) For the block,the final velocity $v = 0$,time $t = 2 \, s$,and displacement $s = 1 \, m$.Using the equation of motion $v = u + at$,we get:$0 = u + a 	imes 2 \Rightarrow a = -u/2$.Using the equation $v^2 - u^2 = 2as$,we get:$0 - u^2 = 2 	imes (-u/2) 	imes 1 \Rightarrow -u^2 = -u \Rightarrow u(u - 1) = 0$.Since $u 
eq 0$,the initial velocity is $u = 1 \, m/s$.The acceleration is $a = -u/2 = -0.5 \, m/s^2$.The force of kinetic friction is $f_k = \mu_k mg = m|a|$.Thus,$\mu_k g = |a| \Rightarrow \mu_k = 0.5 / 10 = 0.05$.In this calculation,we have ignored air resistance and other dissipative forces. In a real-world scenario,these additional forces would contribute to the deceleration,meaning the friction force required to stop the box in $2 \, s$ would be less than the value calculated solely based on kinetic friction. Therefore,$\mu_k$ must be less than $0.05$.

$A$ small box resting on one edge of the table is struck in such a way that it slides up to the other edge,$1 \, m$ away,after $2 \, s$. The coefficient of kinetic friction between the box and the table

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