A horizontal force of 10 N is applied on a b | vedclass

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(C) Given: Force $F = 10 \ N$, mass $m = 1.5 \ kg$, coefficient of kinetic friction $\mu_k = 0.2$, $g = 10 \ m/s^2$, initial velocity $u = 0$, time $t

Vedclass · Accepted Answer

$840$

Vedclass · Answer

(C) Given: Force $F = 10 \ N$, mass $m = 1.5 \ kg$, coefficient of kinetic friction $\mu_k = 0.2$, $g = 10 \ m/s^2$, initial velocity $u = 0$, time $t = 6 \ s$.First, calculate the kinetic friction force: $f_k = \mu_k \cdot m \cdot g = 0.2 	imes 1.5 	imes 10 = 3 \ N$.The net force acting on the block is $F_{net} = F - f_k = 10 - 3 = 7 \ N$.The acceleration of the block is $a = F_{net} / m = 7 / 1.5 = 14 / 3 \ m/s^2$.The displacement $s$ in $t = 6 \ s$ is given by $s = ut + (1/2)at^2 = 0 + (1/2) 	imes (14/3) 	imes (6)^2 = (1/2) 	imes (14/3) 	imes 36 = 7 	imes 12 = 84 \ m$.The work done by the applied force is $W = F 	imes s = 10 	imes 84 = 840 \ J$.

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