A weightless spring with spring constant k1 hangs from n elevated support. From

Question

A weightless spring with spring constant k1 hangs from n elevated support. From its lower end hangs a second weightless spring with spring constant k2. An object of mass M hangs from the bottom of the second spring.

a. Find the effective spring constant of the pair of springs as a system.
b. Find the total distance the pair of springs is extended.

c. A certain uniform spring with spring constant k is cut in half. What is the relationship between k and the resulting spring constant k' for each of the resulting smaller springs?

Please be clear as possible and include a picture of graphs or free body diagrams (if you can).
Thanks for your help.

Explanation / Answer

We know that the equation of the force is, F = k x -------------------- (1) By Newton's 2nd law, F = m a = m g -------------------- (2) so from 1 and 2 equations we get m g = k x x = m g / k The total extension distance of the pair fosprings is the sum of the individual extensions given by (b) x = x1 + x2 = (m g / k1) + (m g/ k1) = m g [(1 / k1)+ (1 / k2)] (a) The effective spring constant is k = [(1 / k1) + (1 /k2)] (c) Relation of force applied on spring (F), the increase in length (x) and the force constant (k) is given by F=k.x so k=F/x Now if the spring is cut into half its increase in length will also be half since increase in length is directly proportional to the length of the spring. Now x'=x/2 and the force constant is k'. If same force is applied the relation will be F=k'.x' => F=k'.x/2 => k'=2.F/x => k'=2.k So the force constant of the spring is doubled.

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