Question

The figure above shows the same ideal spring in three different situations.
First without a mass hanging the spring is unstretched.
Next with the blue mass hanging the spring stretches.
Finally with the red mass hanging the spring stretches twice as much as it did with the blue mass.
How much does the red mass weigh compared to the blue mass?

• A. The red mass weighs twice as much as the blue mass.
• B. The red mass weighs four times as much as the blue mass.
• C. The red mass weighs half as much as the blue mass.
• D. The red mass weighs about 1.4 as much as the blue mass.
• E. We cannot determine how much the red mass weighs compared to the blue mass, because we do not know the unstretched length of the spring.

Answer 1

The correct option is A The red mass weighs twice as much as the blue mass.

In equilibrium position the restoring force set up in the spring is equal to the weight of block i.e. $ky=mg$ ,

where $y=$ stretch in spring , $k=$spring constant ,

now as both the springs are not moving in any direction so they are in equilibrium position ,

for blue mass $m_{B}g=k{y}_B$ ,

for red mass $m_{R}g=k{y}_R$ , we are using same springs in both positions so k will be constant ,

but given that , $2y_B=y_R$ ,

so $m_{R}g=k\times 2y_B=2k{y}_B$

therefore $m_R/m_B=2$ ,

or $m_R=2m_B$