Question 2.13: A truck weighing 2500 kgf and moving at a speed of 2.5 m/s h...

We know that stiffness of spring, k, is given by

k=\frac{G d^4}{64 R^3 n}

Putting the values of G, d, R and n, we get

k=\frac{1 \times 10^6 \times(25 / 10)^4}{64 \times(10)^3 \times 25}=24.41  \text{kgf/cm}

From the condition given in the problem, change in kinetic energy is equal to the overall strain energy of the spring. Mathematically,

\frac{1}{2} m ν^2=\left\lgroup\frac{1}{2} k \delta^2 \right\rgroup \times N

where N is the number of springs.

\begin{aligned} N & =\frac{W}{g} ν^2 \times \frac{1}{k \delta^2} \\ & =\frac{W}{g} \times \frac{ν^2}{k \delta^2} \\ & =\frac{2500}{981} \times \frac{(2.5 \times 100)^2}{24.41 \times(20)^2} \\ & =16.3 \cong 17 \end{aligned}

Thus, the required number of turns is 17.