A block of mass 0.5 kg is at rest on a horiz | vedclass

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(C) Given: Applied force $F_{	ext{app}} = 5 \ N$, mass $m = 0.5 \ kg$, coefficient of kinetic friction $\mu_k = 0.2$, time $t = 4 \ s$, initial veloc

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$256$

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(C) Given: Applied force $F_{	ext{app}} = 5 \ N$, mass $m = 0.5 \ kg$, coefficient of kinetic friction $\mu_k = 0.2$, time $t = 4 \ s$, initial velocity $u = 0$, acceleration due to gravity $g = 10 \ m/s^2$.The kinetic frictional force is given by $f_k = \mu_k N = \mu_k mg$.$f_k = 0.2 	imes 0.5 	imes 10 = 1 \ N$.The net force acting on the block is $F_{	ext{net}} = F_{	ext{app}} - f_k = 5 - 1 = 4 \ N$.The acceleration of the block is $a = \frac{F_{	ext{net}}}{m} = \frac{4}{0.5} = 8 \ m/s^2$.The velocity of the block after $4 \ s$ is $v = u + at = 0 + (8 	imes 4) = 32 \ m/s$.The kinetic energy of the block is $K.E. = \frac{1}{2} mv^2 = \frac{1}{2} 	imes 0.5 	imes (32)^2 = 0.25 	imes 1024 = 256 \ J$.

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