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Is the amount of stretch of the springs proportional to the hanging mass? Explain briefly.



Derive a theoretical expression to find k-equivalent for springs in parallel and springs in series.

Sagot :

Manishagiri1199idn

Answer:

When connected in parallel, the extension of both springs is the same, and the total elastic force will be equal to the sum of the forces in each spring: x=x1=x2,F=F1+F2. F=F1+F2,⇒kx=k1x1+k2x2=(k1+k2)x,⇒k=k1+k2.

Yes. The amount of stretch of a spring is proportional to the hanging mass.

According to Hooke's Law, the restoring force of the spring F is:

                   F = -kx

          k is the spring constant and x is the stretch of spring.

The restoring force F is in opposite direction of the weight of the hanging mass which is mg

                  ⇒ F = -mg

                     mg = -kx

                     x = mg/k

Hence x, the stretch of the spring is directly proportional to the hanging mass.

(i) If two springs are connected in parallel, their k- equialent is

                     [tex]k= k_{1}+k_{2}[/tex]

(ii) If two springs are connected in series, their k- equialent is

                    [tex]\frac{1}{k}=\frac{1}{k_{1} } +\frac{1}{k_{2} }[/tex]

Learn more about Hooke's Law:

https://brainly.com/question/2077015

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