In the arrangement shown in figure, string is light and inextensible and friction is absent every where. The speed of both blocks after the block `'A'` has ascend a height of `1m` will be `(g=10 m//s^(2))`

Explanation

Friction is absent. Therefore, mechanical energy of the system will remain conserved. From constraint relations we see that speed of both the blocks will be same. Suppose it is v. Here gravitational potential energy of 2 kg block is decreasing while gravitational potential energy of 1 kg block is increasing.
Similarly, kinetic energy of both the blocks is also increasing.
`:. m_(B)gh=m_(A)gh+(1)/(2)m_(A)v^(2)+(1)/(2)m_(B)v^(2)`
or `" " (2)(10)(1)=(1)(10)(1)+(1)/(2)(1)v^(2)+(1)/(2)(2)v^(2)`
or `" " 20=10+0.5 v^(2)+v^(2) " or "1.5 v^(2)=10`
`:. v^(2)=6.67 m^(2)s^(-2) " or " v=2.58 ms^(-1)`