A block is projected with speed 20 ,m/s on a | vedclass

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(A) The retardation $a$ of the block is given by $a = \frac{dv}{dt} = -\mu g$.Integrating this expression,we get $\Delta v = v_f - v_i = -g \int_{0}^{

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(A) The retardation $a$ of the block is given by $a = \frac{dv}{dt} = -\mu g$.Integrating this expression,we get $\Delta v = v_f - v_i = -g \int_{0}^{4} \mu \,dt$.The term $\int_{0}^{4} \mu \,dt$ represents the area under the $\mu-t$ curve from $t = 0$ to $t = 4 \,s$.The area consists of a rectangle from $t = 0$ to $2 \,s$ and a trapezoid from $t = 2$ to $4 \,s$.Area $= (2 	imes 0.5) + \frac{1}{2} 	imes (0.5 + 0.3) 	imes (4 - 2) = 1 + 0.8 = 1.8$.Now,substituting the values into the velocity equation:$v_f - 20 = -10 	imes (1.8) = -18$.$v_f = 20 - 18 = 2 \,m/s$.

$A$ block is projected with speed $20 \,m/s$ on a rough horizontal surface. The coefficient of friction $(\mu)$ between the surfaces varies with time $(t)$ as shown in the figure. The speed of the body at the end of $4 \,s$ will be ............ $m/s$ $(g = 10 \,m/s^2)$.

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