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puteri [66]
3 years ago
6

Consider two mass-less springs connected in series. Spring 1 has a spring constant k₁, and spring 2 has a spring constant k₂. A

constant force of magnitude F is being applied to the right. When the two springs are connected in this way, they form a system equivalent to a single spring of spring constant k.
What is the effective spring constant k of the two-spring system? Express the effective spring constant in terms of k₁ and k₂.

Physics
1 answer:
olga2289 [7]3 years ago
6 0

Explanation:

Below is an attachment containing the solution.

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Answer:

The correct option is d: T₁ > T₂ > T₃.  

Explanation:

Let's evaluate each tension.

<u>Case T₃.</u>

T_{3} - W_{3} = 0

For the system to be in equilibrium, the algebraic sum of the tension force (T) and the weight (W) must be equal to zero. The minus sign of W is because it is in the opposite direction of T.          

T_{3} = W_{3}          

Since W₃ = mg, where <em>m</em> is for mass and <em>g</em> is for the acceleration due to gravity, we have:                

T_{3} = W_{3} = mg  (1)                                                                                                     <u>Case T₂.</u>

T_{2} - (T_{3} + W_{2}) = 0    

T_{2} = T_{3} + W_{2}   (2)

By entering W₂ = 2mg and equation (1) into eq (2) we have:

T_{2} = T_{3} + W_{2} = mg + 2mg = 3mg

<u>Case T₁.</u>

T_{1} - (T_{2} + W_{1}) = 0  

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Knowing that W₁ = 3mg and T₂ = 3mg, eq (3) is:

T_{1} = 3mg + 3mg = 6mg        

Therefore, the correct option is d: T₁ > T₂ > T₃.  

I hope it helps you!

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