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(A) The body moves with a uniform velocity,which means the applied force is equal to the force of kinetic friction.Force of friction $f_k = \mu_k N =

Vedclass · Accepted Answer

$9.33$

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(A) The body moves with a uniform velocity,which means the applied force is equal to the force of kinetic friction.Force of friction $f_k = \mu_k N = \mu_k mg$.Given: $m = 2\, kg$,$\mu_k = 0.20$,$v = 2\, m/s$,$t = 5\, s$,$g = 9.8\, m/s^2$,and $J = 4.2\, J/cal$.Distance covered $d = v 	imes t = 2\, m/s 	imes 5\, s = 10\, m$.Work done against friction $W = f_k 	imes d = (\mu_k mg) 	imes d$.$W = 0.20 	imes 2\, kg 	imes 9.8\, m/s^2 	imes 10\, m = 39.2\, J$.Heat generated $Q = \frac{W}{J} = \frac{39.2\, J}{4.2\, J/cal} \approx 9.33\, cal$.

$A$ body of mass $2\, kg$ is being dragged with uniform velocity of $2\, m/s$ on a rough horizontal plane. The coefficient of friction between the body and the surface is $0.20$. The amount of heat generated in $5\, s$ is ........ $cal$ $(J = 4.2\, J/cal$ and $g = 9.8\, m/s^2)$

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