(C) Given mass $m = 2\,kg$,initial velocity $u = 4\,ms^{-1}$ at $x = 0.5\,m$.The retarding force is $F = -kx = -12x$.Using Newton's second law,$F = ma

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(C) Given mass $m = 2\,kg$,initial velocity $u = 4\,ms^{-1}$ at $x = 0.5\,m$.The retarding force is $F = -kx = -12x$.Using Newton's second law,$F = ma$,so $a = \frac{F}{m} = \frac{-12x}{2} = -6x$.We know $a = v \frac{dv}{dx}$,so $v \frac{dv}{dx} = -6x$.Integrating both sides: $\int_{4}^{v} v \, dv = \int_{0.5}^{1.5} -6x \, dx$.$\left[ \frac{v^2}{2} \right]_{4}^{v} = -6 \left[ \frac{x^2}{2} \right]_{0.5}^{1.5}$.$\frac{v^2 - 16}{2} = -3 [ (1.5)^2 - (0.5)^2 ]$.$\frac{v^2 - 16}{2} = -3 [ 2.25 - 0.25 ] = -3 [ 2 ] = -6$.$v^2 - 16 = -12$.$v^2 = 4$,which gives $v = 2\,ms^{-1}$.

Class 11 Physics Work, Energy, Power and Collision Work Done by Variable Force and Force-Displacement Graph

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$A$ block of mass $2\,kg$ moving on a horizontal surface with speed of $4\,ms^{-1}$ enters a rough surface ranging from $x = 0.5\,m$ to $x = 1.5\,m$. The retarding force in this range of rough surface is related to distance by $F = -kx$ where $k = 12\,Nm^{-1}$. The speed of the block as it just crosses the rough surface will be ........... $ms^{-1}$.

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Work Done by Variable Force and Force-Displacement Graph

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