Answered

Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

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 :

manishagiri1199

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

We appreciate your time. Please come back anytime for the latest information and answers to your questions. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.