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  • Both springs have a constant of 25N/m and the block is motionless. If the bottom spring is compressed 0.4m past its equilibrium and the block has a mass of 3kg, how far is the top spring stretched

Both springs have a constant of 25N/m, and the block is motionless. If the bottom spring is compressed 0.4m past its equilibrium and the block has a mass of 3kg, how far is the top spring stretched past its equilibrium? (answer in meters)

 

Answers (1)

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Given- 

  • Both springs have a spring constant of 25 N/m
  • Bottom spring is compressed by 0.4 meters
  • The mass of the block is 3 kg
  • Block is motionless (in equilibrium)

To find- How far the top spring is stretched = x

Solution- Force from bottom spring (upward) = spring constant × compression = 25 × 0.4 = 10 N

Weight of the block (downward) = mass × gravity = 3 × 9.8 = 29.4 N

Force from top spring (downward) = 25 x

At equilibrium, Upward force = Downward force

So, 10 = 25x + 29.4

x= -0.776 meters (The negative just means the spring is pulling down, which is expected)

Posted by

Saniya Khatri

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