The diagram below represents the graph of the force applied in stretching a spiral spring against the corresponding extension produced within its elastic limit.
Using the notations on the graph, determine the:
(a) force constant of the spring;
(b) work done in stretching the spring from 10 x 10\(^{-2}\)m to 20 x 10\(^{-2}\)m.
a) Force constant of the spring = slope of the graph = \(\frac{\Delta force}{\Delta Extension}\)
= \(\frac{60 — 0}{20 - 0 \times 10^{-2}} = \frac{60}{20 \times 10^{-2}}\)
= 300Nm\(^{-2}\)
(b) Work done = Area under the trapezium ABCD
= 1/2 (30 + 60) x (20 — 10) x 10\(^{-2}\) = 4.5J
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}