A 20 kgblock on a horizontal surface is attached to a horizontal spring of spring constant k = 4.0 kN/m. The block is pulled to the right so that the spring is stretched 10 cmbeyond its relaxed length, and the block is then released from rest. The frictional force between the sliding block and the surface has a magnitude of 80 N. (a) What is the kinetic energy of the block when it has moved 2.0 cmfrom its point of release? (b) What is the kinetic energy of the block when it first slides back through the point at which the spring is relaxed? (c) What is the maximum kinetic energy attained by the block as it slides from its point of release to the point at which the spring is relaxed?

The mass of the block is, m = 20 kg

The spring constant of the spring is,$\mathrm{k}=4.0\mathrm{kN}/\mathrm{m}\left(\frac{{10}^3\mathrm{N}/\mathrm{m}}{1\mathrm{kN}/\mathrm{m}}\right)=4.0\times {10}^3\mathrm{N}/\mathrm{m}$

The stretched length of the block is,$\times =10\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.10\mathrm{m}$

The magnitude of the force is,$\mathrm{F}=80\mathrm{N}$

We can use the law of conservation of energy to solve this. It states that the sum of potential and kinetic energy is equal to mechanical energy and this always remains constant.

Formulae:

The kinetic energy of a body while stretching, $∆\mathrm{K}={\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}=-\mathrm{kxdx}$ (1)

Using equation (1), we can get the change in the kinetic energy of the block as given:

$$\begin{gathered} ∆\mathrm{K}={\int }_{0.10}^{0.08}-\mathrm{kxdx} \\ =-0.5\times \mathrm{k}\times {\left[{\mathrm{x}}^2\right]}_{0.01}^{0.08} \\ =-0.5\times 4.0\times {10}^3\mathrm{N}/\mathrm{m}\times \left({\left(0.08\mathrm{m}\right)}^2-{\left(0.1\mathrm{m}\right)}^2\right) \\ =-7.2\mathrm{J} \end{gathered}$$

Now, applying the law of conservation of energy, we can get the kinetic energy of the block as:

$$\begin{gathered} \mathrm{K}+∆\mathrm{K}=-{\mathrm{f}}_{\mathrm{r}}\times \mathrm{x} \\ \mathrm{K}-7.2\mathrm{J}=-80\mathrm{N}\times 0.02\mathrm{m} \\ \mathrm{K}=5.6\mathrm{J} \end{gathered}$$

Hence, the value of the energy is 5.6 J.

Using equation (1), the kinetic energy of the block when it slides back to that point is given as:

Hence, the kinetic energy of the block is 12 J.

The net displacement is given by:

$$\begin{gathered} \mathrm{x}=0.10\mathrm{m}-\mathrm{d} \\ 0.02\mathrm{m}=0.10\mathrm{m}-\mathrm{d} \\ \mathrm{d}=0.08\mathrm{m} \end{gathered}$$

The change in kinetic energy is given using equation (i) as follows:

$$\begin{gathered} ∆\mathrm{K}={\int }_{0.10}^{0.02}-\mathrm{kxdx} \\ =-0.5\times \mathrm{k}\times {\left[{\mathrm{x}}^2\right]}_{0.01}^{0.02} \\ =-0.5\times 4\times {10}^3\mathrm{N}/\mathrm{m}\times \left[{\left(0.02\mathrm{m}\right)}^2-{\left(0.01\mathrm{m}\right)}^2\right] \\ =-19.2\mathrm{J} \end{gathered}$$

Now applying the law of conservation of energy, the maximum energy obtained by the block is given as:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{ma}}+∆\mathrm{K}=-{\mathrm{f}}_{\mathrm{r}}\times \mathrm{d} \\ {\mathrm{K}}_{\mathrm{ma}}-19.2\mathrm{J}=-80\mathrm{N}\times 0.08\mathrm{m} \\ {\mathrm{K}}_{\mathrm{ma}}=12.8\mathrm{J} \end{gathered}$$

Hence, the value of the maximum energy is 12.8 J.