Question

The system 8.154 shown in Fig. is released from rest with mass 2 kg in contact with the ground. Pulley and spring are mass less, and friction is absent everywhere. The speed of 5 kg block when 2 kg block leaves the contact with the ground is (force constant of the spring k = 40 $N{m}^{-1}$ and g = 10 m$s^{-2}$) :

• A. $\sqrt{2}m{s}^{-1}$
• B. $2\sqrt{2}m{s}^{-1}$
• C. $2m{s}^{-1}$
• D. $3\sqrt{2}m{s}^{-1}$

Answer 1

The correct option is B $2\sqrt{2}m{s}^{-1}$

let x be the extension of the spring at rest of 2kgblock

Kx = 2g

x = $\frac{2g}{K}=\frac{20}{40}=\frac{1}{2}m$

from law of conservation of energy

$mgx=\frac{1}{2}K{x}^2=\frac{1}{2}m{v}^2$

$\frac{1}{2}m{v}^2=mgx-\frac{1}{2}K{x}^2$

$v^2=2gx-\frac{K{x}^2}{m}$ = $2\left(10\right)\frac{1}{2}$- $\frac{40}{4\left(5\right)}=8$

$v^2=8⟹v=2\sqrt{2}\frac{m}{s}$