Question

A spring of force constant 900 Nm^-1 is cut into two pails, keeping ratio of lengths of the parts 3 : 2. Both the spring parts are connected horizontally to a block of mass 2kg, placed on a horizontal frictionless surface. Initially the spring parts are non deformed as shown. The block is then pushed to the left by 5cm and released. The acceleration of the block immediately after the release is:

• A. 25 m/s²
• B. 9 m/s²
• C. 30 m/s²
• D. 37.50 m/s²

Answer 1

Answer: (D)

37.50 m/s²

Answer 2

If initial length of the spring is/then using kl = k₁l₁ = k₂l₂ for parts of the spring

⇒ k₁ = $\frac { \mathrm { kl } } { l _ { 1 } } = 900 \cdot \frac { 1 } { ( 31 / 5 ) } = 1500 \mathrm { Nm } ^ { - 1 }$ k₂ =

$\frac { \mathrm { kl } } { 1 _ { 2 } } = 900 \cdot \frac { 1 } { ( 21 / 5 ) } = 2500 \mathrm { Nm } ^ { - 1 }$ As spring are connected in parallel, effective spring constant

k_eq = k₁ + k₂ = 3750 Nm^-1

Net force on the block at the moment of release in horizontal direction

= K_eq. x

= (3750) × (5/100)

Acceleration = F_net/m = $\frac { 3750 } { 5 } \times \left( \frac { 5 } { 100 } \right)$ = 37.50 ms^-2