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  • Consider the following system: <img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAGEAAAB/CAYAAAD/2qxIAAAgAElEQVR4nOy7Z7dtV3nn6e/QNco9qttRBiQhIYLIwSIbMDmVTTBlY7ApKMcyTgQDFhZC4YYT7sl5

Consider the following system:

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)

 

Option: 1

0.8


Option: 2

0.4


Option: 3

0.6

 


Option: 4

1


Answers (1)

best_answer

Solution :
Combination of Spring:-

  • Series combination:- 

\frac{1}{k_{eq}}=\frac{1}{k_{1}}+\frac{1}{k_{2}}

  • Parallel combination:-

k_{eq}=k_{1}+k_{2}

By using this concept, let us solve this problem - 

Since the block is motionless, we know that our forces will cancel out:

\therefore F_{net}=0

There are three forces in play: one from each spring, as well as the force of gravity. If we assume that forces pointing up are positive, we can write:

F_{spring.top}+F_{spring.bot}-mg = 0

Plugging in expressions for each spring force, we get:

kx_{top}+kx_{bot}-mg = 0

Rearranging for the displacement of the top spring, we get:

x_{top}=\frac{mg-kx_{bot}}{k}=\frac{(3kg)(10{{m/s^{2}}})-(25{N/m})(0.4m)}{25{N/m}}=0.8m

 

Posted by

manish painkra

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