Question

Figure shows two blocks A and B, each having a 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. The 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/$s^2$.

• A. 1.5 m/s
• B. 2 m/s
• C. 1 m/s
• D. None of these

Answer 1

The correct option is A

1.5 m/s

m = 320g = 0.32 kg

K = 40 N/m

H = 40cm = 0.4m

G = 10 m/$s^2$

From the free body diagram,

kx cos$\theta$ = mg

(When the block breaks off R = 0)

$\Rightarrow cos\theta =\frac{mg}{kx}$

So, $\frac{0.4}{0.4+x}=\frac{3.2}{40x}\Rightarrow 16x=3.2x+1.28\Rightarrow x=0.1m$

So, s = AB = $\sqrt{(h+x{)}^2-h^2}=\sqrt{(0.5{)}^2-(0.4{)}^2}=0.3m$

Let the velocity of the body at B be v

Charge in K.E = work done (for the system)

$(\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2)=\frac{-1}{2}k{x}^2+mgs$

$\Rightarrow \left(0.32\right)\times v^2=-\left(\frac{1}{2}\right)\times 40\times (0.1{)}^2+0.32\times 10\times (0.3)\Rightarrow v=1.5m/s$.