Question

Figure shows two blocks A and B, each having a mass of 320 g connected by a light string passing over a smooth and 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/s².
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

Given in the question :-
Mass of the block = 320 gm = 0.32 Kg
g = 10m/s²h = 40cm = 0.4m
k = 40N/m.

Kinetic energy at Initial system = 0 
Refer to The attachment , Let spring make an angle θ  and Force due to the spring T , If the velocity is v.
Then kinetic energy on the block = 2 × 1/2 (mv²) =mv²
Now change in K.E = mv² -0 = mv² .

Elongation in the spring , 
T = kx
Where x = hcosecθ - h .
T = k(hcosecθ -h)

By equating vertical forces on the block , 
T.sinθ =mg
T = mg/sinθ
k(h.cosecθ -h) =mg/sinθ .
sinθcosecθ -sinθ =mg/kh
1-sinθ =mg/kh
sinθ = 1-mg/kh
sinθ = 1- (0.32 × 10) / (0.4× 40)
sinθ = 1- 3.2/16
sinθ = 1-0.2
sinθ = 0.8.

Now elongation x,
x = hcosecθ -h
x = h/sinθ -h
x =0.4/0.8 -0.4
x= 0.1 m.

Now for displacement s,
s= hcotθ .
s= 0.4 × 3/4   [cotθ =cosθ/sinθ]  
s = 0.3 m

Therefore Change in Kinetic energy = Work done by all forces
mg² = -1/2 kx²+ mgs    [-ve sign here since elongation is in opp. direction ]
v² = gs - 1/2 kx²/m
v² = 10 × 0.3 - 1/2 (40 × 0.1²) /0.32 .
v² = 3 -  0.2 /0.32
v² = 3 -0.625
v² = 2.375 
v = √2.375
v = 1.54 m/s.

Hence the velocity of the block A is 1.54 m/s.


Hope it Helps.