Question

The system shown in the figure is in equilibrium. Masses m₁ and m₂ are 2 kg and 8kg respectively. Spring constants k₁ and k₂ are 50 N/m and 70 N/m respectively. If the compression in second spring is 0.5 m. What is the compression in first spring ? (Both springs have the same natural length) –

• A. 1.3m
• B. –0.5m
• C. 0.5m
• D. 0.9m

Answer 1

Answer: (B)

–0.5m

Answer 2

As the springs are fixed to the horizontal and have the same natural length, hence if one spring is compressed, the other must be expanded. Hence, the compression will be negative.

F.B.D. of m₂ T + F₂ = 80 N

and F₂ = 70 × 0.5 = 35 N

\ T = 80 – 35 = 45 N

F.B.D. of m_l T + F₁ + mg

or 45 = F₁ + 20

or F₁ = 25N

\ x₁ = $\frac { 25 } { \mathrm { k } _ { 1 } } = \frac { 25 } { 50 }$ = 0.5m

\ Compression in first spring = – 0.5 m