A block of mass 10, kg moving at 10, m/s is r | vedclass

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(B) Given: Mass $m = 10\, kg$,initial velocity $u = 10\, m/s$,final velocity $v = 0\, m/s$,coefficient of friction $\mu = 0.2$,and acceleration due to

Vedclass · Accepted Answer

$25$

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(B) Given: Mass $m = 10\, kg$,initial velocity $u = 10\, m/s$,final velocity $v = 0\, m/s$,coefficient of friction $\mu = 0.2$,and acceleration due to gravity $g = 10\, m/s^2$.According to Newton's second law,the frictional force is $f = \mu N = \mu mg$.The retardation $a$ produced by this force is $a = \frac{f}{m} = \mu g$.Substituting the values: $a = 0.2 	imes 10 = 2\, m/s^2$.Using the third equation of motion $v^2 = u^2 - 2as$:$0^2 = 10^2 - 2 	imes 2 	imes s$$0 = 100 - 4s$$4s = 100$$s = 25\, m$.Therefore,the block will stop after travelling a distance of $25\, m$.

$A$ block of mass $10\, kg$ moving at $10\, m/s$ is released to slide on a rough surface having a coefficient of friction $0.2$. It will stop after travelling a distance of ........ $m$.

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