Answered

Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Discover comprehensive answers to your questions from knowledgeable professionals on our user-friendly platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

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

Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.