A body of mass 1 ,kg is moving with a velocit | vedclass

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(A) Given: Mass $m = 1 \,kg$, initial velocity $u = 10 \,ms^{-1}$, final velocity $v = 0 \,ms^{-1}$, coefficient of kinetic friction $\mu = 0.4$, and

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(A) Given: Mass $m = 1 \,kg$, initial velocity $u = 10 \,ms^{-1}$, final velocity $v = 0 \,ms^{-1}$, coefficient of kinetic friction $\mu = 0.4$, and $g = 10 \,ms^{-2}$.When the force is removed, the only horizontal force acting on the body is the kinetic frictional force $f_k = \mu N = \mu mg$.According to Newton's second law, the retardation $a$ is given by $a = \frac{f_k}{m} = \frac{\mu mg}{m} = \mu g$.Substituting the values: $a = 0.4 	imes 10 = 4 \,ms^{-2}$.Using the first equation of motion $v = u - at$ (where $a$ is retardation):$0 = 10 - 4t$$4t = 10$$t = \frac{10}{4} = 2.5 \,s$.Thus, the body comes to rest in $2.5 \,s$.

$A$ body of mass $1 \,kg$ is moving with a velocity $10 \,ms^{-1}$ on a horizontal rough surface having a coefficient of kinetic friction $0.4$. If the constant force is removed, the body comes to rest in a time (Acceleration due to gravity $g = 10 \,ms^{-2}$). (in $\,s$)

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