Answer to Question #134129 in Mechanics | Relativity for adeola adegoke

Question

the length of a spring is spring is 16cm when a load of 5n is attached to the spring is 20n the length is 19cm .calculate th
(i) force constant of the spring
(ii)original length of spring

Answer 1

Explanations & Calculations

• Take the force constant as $\small k$ and the original length as $\small l_0$.
• Then considering the extensions & the equation : $\small F=kx$, followings can be written to calculate the needed data.

$\qquad\qquad \begin{aligned} \small l_0+e_1 &= \small 0.16m \,\,\,\,\to \to \,\,\,\ e_1=0.16-l_0\\ \small ke_1 &= \small k(0.16-l_0)=5N\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdots\cdots(1)\\ \\ \small l_0+e_2 &= \small 0.19m \,\,\,\,\to \to \,\,\,\,e_2=0.19-l_0\\ \small ke_2 &= \small k(0.19-l_0 ) =20N \,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdots\cdots(2) \end{aligned}$

• By (2) - (1),

$\qquad\qquad \begin{aligned} \small 15N &= \small k(0.03m)\\ \small \bold{k} &= \small \bold{500Nm^{-1}} \end{aligned}$

• From equation (1),

$\qquad\qquad \begin{aligned} \small 500Nm^{-1} \times (0.16-l_0)&= \small 5N\\ \small \bold{l_0} &= \small \bold{0.15m} = \bold{15cm} \end{aligned}$