Question

Figure (8-E12) shows two blocks A and B, each of mass of 320 g connected by a light string passing over a smooth light pulley. The horizontal surface on which the block A can slide is smooth. Block A is attached to a spring of spring constant 40 N/m whose other end is fixed to a support 40 cm above the horizontal surface. Initially, the spring is vertical and unstretched when the system is released to move. Find the velocity of the block A at the instant it breaks off the surface below it. Take g = 10 m/s2.

Answer 1

$\huge{\boxed{\mathcal\pink{\fcolorbox{red}{yellow}{Answer}}}}$

1.54 m/s

Answer 2

The velocity of the block A at the instant it breaks off the surface below it is 1.54 m/s.

Explanation:

Block's mass = 320 gm = 0.32 kg                                                  

                    g = 10 m/s²

                    h = 40 cm = 0.4 m

Initial kinetic energy = 0

Let spring make an angle θ and force because of the T spring

Where, velocity is v.

Block's kinetic energy = 2 × $\frac{1}{2}$ (mv²) = mv²

Now, change in kinetic energy = mv² - 0 = mv²

In the spring elongation,

                                        T = kx

Where, x = h cosec θ   - h

            T = k ( h cosec θ - h )

By the vertical forces of the block,            

                         T . sin θ = mg

                                     $\mathrm{T}=\frac{m g}{\sin \theta}$

                $\mathrm{K}(\mathrm{h} \cdot \csc \theta-\mathrm{h})=\frac{m g}{\sin \theta}$

     sin θ . cosec θ  - sin θ  = $\frac{mg}{kh}$

                            1 - sin θ   = $\frac{mg}{kh}$

                                 sin θ  = 1 - $\frac{mg}{kh}$

                                 sin θ  = 1 - $\frac{(0.32 . 10)}{(0.4.40)}$      

                                 sin θ  = 1 -$\frac{3.2}{16}$

                                 sin θ  = 1 - 0.2

                                  sin θ  = 0.8

Now elongation x,

                              x = h cosecθ - h

                              x = $x=\frac{h}{\sin \theta}-h$

                              x = $\frac{0.4}{0.8} - 0.4$

                              x = 0.5 - 0.4

                              x = 0.1 m

Now, for displacement s,

                              s = h cot θ  

                              s = 0.4 × $\frac{3}{4}$    ( $\cot \theta=\frac{\cos \theta}{\sin \theta}$  )

                              s = 0.3 m

Therefore, change in kinetic energy = work done by all forces

                          mg² = - $\frac{1}{2}$ kx²  + mgs

[ -ve sign here indicates elongation is in opposite direction ]

                              v² = gs - $\frac{1}{2 m}$ kx²

                              v² = 10 × 0.3 - $\frac{1}{(2).(0.32)}$  (40×0.1²)

                              v² = 3 - $\frac{0.2}{0.32}$

                               v² = 3 - 0.625

                               v² = 2.375

                                v = $\sqrt{2.375}$

                               v = 1.54 m/s

Whereas, block A velocity is 1.54 m/s.

Therefore, the velocity of block A is 1.54 m/s.