Question

A spring of free length 240 mm is 200 mm long when carrying a compressive load of 50 N. When fully compressed, the spring is 110 mm long. Calculate the spring rate and the work done in compressing the spring from 175 mm long to 120 mm in length. 

Clearly show these values in your calculation

k=?

F1=?

F2=?

ave F=?

W=?

Answer 1

$\Delta x = 240-200 = 40 (mm)=0.04(m)$

$F= 50N$

$$k = \frac{F}{\Delta x}$$

$$k = \frac{50\ N}{ 0.04\ m} =1250 \ N/m$$

$$F_1=1250\ N/m\cdot(0.24\ m-0.11\ m)=162.5\ N$$

$$F_2=1250\ N/m\cdot(0.175\ m-0.12\ m)=68.75\ N$$

Work done $W$

$$W = \dfrac{k(\Delta x)^2}{2}$$ $$W_2 = \dfrac{1250\ N/m\cdot(0.175\ m-0.12\ m)^2}{2}$$

$$=1.890625\ J$$

$$ave F=\dfrac{1.890625\ J}{0.175\ m-0.12\ m}=34.375\ N$$