A small body of mass 500 ,g moves on a rough | vedclass

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(B) Since the retardation is constant,the average velocity is given by:$v_{av} = \frac{u + v}{2} = \frac{2 + 0}{2} = 1 \,m/s$The frictional force $f$

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(B) Since the retardation is constant,the average velocity is given by:$v_{av} = \frac{u + v}{2} = \frac{2 + 0}{2} = 1 \,m/s$The frictional force $f$ is given by:$f = \mu N = \mu mg$Given:Coefficient of friction,$\mu = 0.3$Mass of body,$m = 500 \,g = 0.5 \,kg$Acceleration due to gravity,$g = 10 \,m/s^2$Substituting the values:$f = 0.3 	imes 0.5 	imes 10 = 1.5 \,N$The average power $P_{av}$ developed by the frictional force is:$P_{av} = f 	imes v_{av}$$P_{av} = 1.5 \,N 	imes 1 \,m/s = 1.5 \,W$Thus,the absolute value of the average power is $1.5 \,W$.

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