Question

As illustrated in the figure, a spring with spring constant k is stretched from x = 0 to x = 3d, where x = 0 is the equilibrium position of the spring. During which interval is the largest amount of energy required to stretch the spring?

Answer 1

Hooke's law states that the force required to stretch or compress a spring to a distance is direclty proportional to displacement.

$F\propto -x$

As the spring extends farer distance from the equilibrium and then it requires more force to stretch the spring.

The force increase linealry as the dispalcement increase from equilibrium position.

The work done on the spring should be stored as elastic potential energy, which is given as,

$U=(1/2)kx^2$

Thus, potential energy depends upon the elongation of spring. Therefore, the potential energy becomes greater when displacment is larger.

Hence, the largest potential required to stretch the spring is from 2d to 3d.

Answer 2

(a) The potential energy stored in a spring mass system is,

$$ U=\frac{1}{2} k x^{2} $$

The energy required is equal to the change in potential energy of the spring. The energy required stretch the spring from \(x=0\) to \(x=d\) is,

$$ \begin{aligned} E &=U_{f}-U_{i} \\ &=\frac{1}{2} k d^{2}-\frac{1}{2} k(0)^{2} \\ &=\frac{1}{2} k d^{2} \end{aligned} $$

(b) The energy required stretch the spring from \(x=d\) to \(x=2 d\) is,

$$ \begin{aligned} E &=U_{f}-U_{i} \\ &=\frac{1}{2} k(2 d)^{2}-\frac{1}{2} k(d)^{2} \\ &=4\left(\frac{1}{2} k d^{2}\right)-\frac{1}{2} k d^{2} \\ &=3\left(\frac{1}{2} k d^{2}\right) \end{aligned} $$

(c) The energy required stretch the spring from \(x=2 d\) to \(x=3 d\) is,

$$ \begin{aligned} E &=U_{f}-U_{i} \\ &=\frac{1}{2} k(3 d)^{2}-\frac{1}{2} k(2 d)^{2} \\ &=9\left(\frac{1}{2} k d^{2}\right)-4\left(\frac{1}{2} k d^{2}\right) \\ &=5\left(\frac{1}{2} k d^{2}\right) \end{aligned} $$

Thus, the energy required is more during the stretch from \(x=2 d\) to \(x=3 d\).