Question

A 2 kg block is connected with two springs of force constants k₁ = 100 N/m and k₂ = 300 N/m as shown in figure. The block is released from rest with the springs unstretched. The acceleration of the block in its lowest position is : (g = 10 m/s²) –

• A. Zero
• B. 10 m/s² upwards
• C. 10 m/s² downwards
• D. 5 m/s² upwards

Answer 1

Answer: (B)

10 m/s² upwards

Answer 2

Let x be the maximum displacement of block downwards. Then from conservation of mechanical energy:

decrease in potential energy of 2 kg block = increase in elastic potential energy of both the springs

\ mg x = $\frac { 1 } { 2 }$ (k₁ + k₂) x²

or x = $\frac { 2 \mathrm { mg } } { \mathrm { k } _ { 1 } + \mathrm { k } _ { 2 } }$ = $\frac { ( 2 ) ( 2 ) ( 10 ) } { 100 + 300 }$ = 0.1 m

Acceleration of block in this position is –

a = $\frac { \left( k _ { 1 } + k _ { 2 } \right) x - m g } { m }$ (upwards)

= $\frac { ( 400 ) ( 0.1 ) - ( 2 ) ( 10 ) } { 2 }$ = 10 m/s² (upwards