91Ó°ÊÓ

The mass of the block is\(m\).

The initial displacement of the block is\({x_{\rm{o}}}\).

The initial speed of the block is\({v_{\rm{o}}}\).

The spring constant is k.

Let v be the maximum speed of the block and x be the maximum stretch of the spring.

For the maximum speed, the total mechanical energy is kinetic, and the spring is at its original length so that there is no potential energy in the spring.

The initial total mechanical energy of the systems is equal to the sum of the kinetic energy of the block due to initial speed and the potential energy of the spring due to stretch.

Then the total initial mechanical energy of the system is as follows:

\(E = \frac{1}{2}mv_{\rm{o}}^2 + \frac{1}{2}kx_{\rm{o}}^2\)

Part (a)

For the maximum speed of the block, the total mechanical energy of the system is kinetic energy as the potential energy is zero.

Then,

\(\begin{aligned}\frac{1}{2}m{\left( v \right)^2} &= E\\\frac{1}{2}m{\left( v \right)^2} &= \frac{1}{2}mv_{\rm{o}}^2 + \frac{1}{2}kx_{\rm{o}}^2\\{v^2} &= v_{\rm{o}}^2 + \frac{k}{m}x_{\rm{o}}^2\\v &= \sqrt {v_{\rm{o}}^2 + \frac{k}{m}x_{\rm{o}}^2} \end{aligned}\)

Hence, the maximum speed of the block is \(\sqrt {v_{\rm{o}}^2 + \frac{k}{m}x_{\rm{o}}^2} \).

Part (b)

For the maximum stretch of the block, the total mechanical energy of the system is the potential energy of the spring,asthe kinetic energy of the block is zero.

Then,

\(\begin{aligned}\frac{1}{2}k{x^2} &= E\\\frac{1}{2}k{x^2} &= \frac{1}{2}mv_{\rm{o}}^2 + \frac{1}{2}kx_{\rm{o}}^2\\{x^2} &= \frac{m}{k}v_{\rm{o}}^2 + x_{\rm{o}}^2\\x &= \sqrt {\frac{m}{k}v_{\rm{o}}^2 + x_{\rm{o}}^2} \end{aligned}\)

Hence, the maximum stretch of the spring is \(\sqrt {\frac{m}{k}v_{\rm{o}}^2 + x_{\rm{o}}^2} \).