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]
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