In Fig.8.52, a 3.5 kg block is accelerated from rest by a compressed spring of spring constant 640 N/m. The block leaves the spring at the spring’s relaxed length and then travels over a horizontal floor with a coefficient of kinetic friction ${\mathit{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}$.The frictional force stops the block in distance D = 7.8 m. What are (a) the increase in the thermal energy of the block–floor system (b) the maximum kinetic energy of the block, and (c) the original compression distance of the spring?

The mass of the block,$m=3.5\mathrm{kg}$

The spring constant of the spring,$k=640\mathrm{N}/\mathrm{m}$

The coefficient of friction,${\mathrm{\mu }}_k=0.25$

The distance at which the blocks comes to rest,$D=7.8\mathrm{m}$

The total energy of the block-floor system is conserved during motion. The spring is initially compressed. When the block leaves the spring at its relaxed position, the potential energy is converted to kinetic energy of the block. As the block continues its motion, it loses its energy to the frictional force of the floor as it finally comes to rest.

Formula:

$$\begin{gathered} \mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2 \\ \mathrm{P}.\mathrm{E}\left(\mathrm{spring}\right)=\frac{1}{2}{\mathrm{kx}}^2 \\ {\mathrm{E}}_{\mathrm{th}}={\mathrm{F}}_{\mathrm{f}}\mathrm{x} \end{gathered}$$

The increase in thermal energy of the block-floor system occurs because of the friction between the block and the floor. Hence

$$\begin{gathered} E_{th}=F_{f}x \\ {E}_{th}={\mathrm{\mu }}_{\mathrm{k}}mgD \\ =0.25\times 3.5\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 7.8\mathrm{m} \\ =66.9\mathrm{J} \\ ~67\mathrm{J} \end{gathered}$$

Hence, the increase in thermal energy of block-floor system = 67 J

The block comes to rest after travelling distance D. Hence, the kinetic energy of the block at the start of this motion is completely converted to the thermal energy. So that we write

$E_{th}=KE$

Hence, the maximum kinetic energy of the block = 67 J

When the block leaves the spring, it is the relaxed position of the spring. So the kinetic energy of the block at this moment is potential energy of the compressed spring.

So we write,

$$\begin{gathered} \mathrm{K}.\mathrm{E}\left(\mathrm{block}\right)=\mathrm{P}.\mathrm{E}\left(\mathrm{spring}\right) \\ \frac{1}{2}{\mathrm{kx}}^2=67\mathrm{J} \\ \frac{1}{2}\times 640\mathrm{N}/\mathrm{m}\times {\mathrm{x}}^2=67\mathrm{J} \\ {\mathrm{x}}^2=\frac{67\mathrm{J}\times 2}{640\mathrm{N}/\mathrm{m}} \\ =0.21 \\ \mathrm{x}=0.46\mathrm{m} \\ =46\mathrm{cm} \end{gathered}$$

Thus, the original compression of the spring = 46 cm