Two smooth blocks A of mass $1 \mathrm{kg}$ and B of mass $2 \mathrm{kg}$ are connected by a light string passing over a smooth pulley as shown. The block $B$ is sliding down with a velocity $2 \mathrm{m}/\mathrm{s}. \mathrm{A}$ force $F$ is applied on the block $A,$ so that the block $B$ will reverse its direction of motion after $3 \mathrm{s}$.

F = 14 N

Setup:

• $m_A=1\text{ kg}$ on horizontal. $m_B=2\text{ kg}$ on a smooth $37^\circ$ incline.

• $B$ initially moves down the plane at $u=2\,\text{m s}^{-1}$.

• It reverses after $3\text{ s}$ ⇒ with constant acceleration, $v=0$ at $t=3\text{ s}$:

a=tv−u​=30−2​=−32​ m s−2

(negative along “down the plane”, so $B$ accelerates up.)

Take $g=10\,\text{m s}^{-2}$, $\sin37^\circ=\tfrac{3}{5}$.

Equations along motion:

For $B$ (down positive):

2a=2gsin37∘−T⇒2(−32​)=12−T⇒T=340​ N.

For $A$ (right positive, same magnitude $a=\tfrac{2}{3}$):

a=F−T⇒32​=F−340​⇒F=14 N.