Super problem by Alok

Classical Mechanics Level 2

Consider a spring with negligible mass that has an unstretched length $l_{0} = 8.8 \times 10^{-2} \text{ m}$ . A body of mass $m_{1} =1.5 \times 10^{-1} \text{ kg }$ is suspended from one end of the spring. The other end of the spring is fixed. After a series of oscillations has died down, the new stretched length of the spring is $l = 9.8 \times 10^{-2} \text{ m}$ .

Assuming that the spring satisfies Hooke’s Law when stretched, what is the spring constant?

The answer is 150.

2 solutions

Alok Patel
Dec 7, 2014

Hooke’s Law states that the extension of the spring is proportional to the applied force (in Latin, “Ut Tensio Sic Vis”); the spring constant is the proportionality constant, k=F/l=m {l}g /l-l {0}=(1.5\times10^{-1}kg)(9.8m/s )/9.8\times10^{-2}-8.8\times10^{-2}=150 N/m

Abhishek Chopra
Dec 6, 2014

F=-kx m a=-kx 1.5 10^-1 10=-k -(9.8 10^-2 - 8.8 10^-2) k=150 n/m

Super problem by Alok

The answer is 150.

2 solutions

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