A boy throws a cricket ball from the boundary to the wicket keeper. If the frictional force due to air $\left(f_a\right)$ cannot be ignored, the forces acting on the ball at the position $X$ are represented by

Two blocks $A$ and $B$ of masses $1.5 \mathrm{kg}$ and $0.5 \mathrm{kg}$ respectively are connected by a massless inextensible string passing over a frictionless pulley as shown in the figure. Block A is lifted until block B touches the ground and then block A is released. The initial height of block A is $80 \mathrm{cm}$ when block B just touches the ground. The maximum height reached by block $\mathrm{B}$ from the ground after the block A falls on the ground is

Two blocks $A$ and $B$ of masses $4 \mathrm{kg}$ and $6 \mathrm{kg}$ are as shown in the figure. A horizontal force of $12 \mathrm{N}$ is required to make $A$ slip over $B$. Find the maximum horizontal force $F_B$ that can be applied on $B$ so that both $A$ and $B$ move together. (take $g=10 \mathrm{m} {\mathrm{s}}^{-2}$)

A $30 \mathrm{kg}$ slab B rests on a frictionless floor as shown in the figure. A $10 \mathrm{kg}$ block A rests on top of the slab- B. The coefficients of static and kinetic friction between the block A and the slab B are $0.60$ and $0.40$ respectively. When block-A is acted upon by a horizontal force of $100 \mathrm{N}$, as shown, find the resulting acceleration of the slab- B. $\left(\mathrm{g}=9.8 \mathrm{m} {\mathrm{s}}^{-2}\right)$