• question_answer A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of $x-$axis is applied to the block. he force is given by $\vec{F}=(4-{{x}^{2}})\hat{i}\,N,$where $x$is in meter and the initial position of the block is $x=0.$The maximum kinetic energy of the block between$x=0$and$x=2.0$ m is A)  2.33 J                         B)  8.67 J C)    5.33 J               D)    6.67 J

From work-energy theorem kinetic energy of block at $x=x$is; $K=\int_{0}^{x}{(4-{{x}^{2}})}.dx$ or $K=4x-\frac{{{x}^{3}}}{3}$ For K to be maximum $\frac{dK}{dx}=0$ or $4-{{x}^{2}}=0$ or $x=\pm 2\,m$ At $x=+2\,m,\,\frac{{{d}^{2}}K}{d{{x}^{2}}}$is negative i.e., kinetic energy (K) is maximum. $\therefore$${{K}_{\max }}=(4)(2)-\frac{{{(2)}^{2}}}{3}=\frac{16}{3}J\,=5.33\,J$ Hence, the correction option is (c).