Respuesta :
When the spring is stretched by 15.2 cm = 0.152 m, the spring exerts a restorative force with magnitude (due to Hooke's law)
[tex]F = kx[/tex]
where [tex]k[/tex] is the spring constant. Solve for [tex]k[/tex].
[tex]49.1\,\mathrm N = k (0.152\,\mathrm m) \implies k \approx 323 \dfrac{\rm N}{\rm m}[/tex]
The amount of work required to stretch or compress a spring by [tex]x\,\mathrm m[/tex] from equilibrium length is
[tex]W = \dfrac12 kx^2[/tex]
Then the work needed to stretch the spring by 15.2 cm is
[tex]W_1 = \dfrac12 \left(343\dfrac{\rm N}{\rm m}\right) (0.152\,\mathrm m)^2 \approx 3.73\,\mathrm J[/tex]
and by 15.2 + 13.7 = 28.9 cm is
[tex]W_2 = \dfrac12 \left(343\dfrac{\rm N}{\rm m}\right) (0.289\,\mathrm m)^2 \approx 13.5\,\mathrm J[/tex]
so the work needed to stretch from 15.2 cm to 28.9 cm from equilibrium is
[tex]\Delta W = W_2 - W_1 \approx \boxed{9.76\,\mathrm J}[/tex]
