Question

In Fig. 8-53, a block of mass$\mathit{m}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{kg}$slides head on into a spring of spring constant$\mathit{k}\mathbf{=}\mathbf{320}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{m}$. When the block stops, it has compressed the spring by$\mathbf{7}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{cm}$. The coefficient of kinetic friction between block and floor is 0.25. While the block is in contact with the spring and being brought to rest, what are (a) the work done by the spring force and (b) the increase in thermal energy of the block-floor system? (c) What is the block’s speed just as it reaches the spring?

Short answer

a) The work done by the spring is - 0.90 J

b) The increase in thermal energy of the block-floor system is 0.46 J

c) The block’s speed just as it reaches the spring is 1.0 m/s

Step-by-step solution

Given data

Mass of the block,$m=2.5\mathrm{kg}$

Spring constant,$\mathrm{k}=320\mathrm{N}/\mathrm{m}$

Spring is compressed by distance, x = 7.5 cm

Coefficient of kinetic friction between block and floor,$\mu =0.25$

Understanding the concept of friction

The block with initial kinetic energy strikes the spring and compresses it. This converts the kinetic energy of the block to the potential energy of the spring. But since there is friction between the block and the floor, some energy is lost as thermal energy. Thus, we have to apply the conservation of energy principle to study the system.

Formulae:

The kinetic energy of the body in motion,$KE=\frac{1}{2}m{v}^2$

The potential energy of a body in motion,$PE=\frac{1}{2}k{x}^2$

Thermal energy of the system due to frictional force,$E_{th}\left(W\right)=F_{f}x$

The frictional force acting on a body, $F_f={\mu }_{k}N={\mu }_{k}mg$

Step 3(a): Calculation of the work done by the spring

Using thework energy theorem, the work done by the spring = change in the energy of the spring. Thus, the work done by the spring is given: (negative as it is case of compression)

$$\begin{gathered} W=\frac{1}{2}k{x}^2 \\ W=-\frac{1}{2}\times 320\times {\left(0.075\right)}^2 \\ =-0.90\mathrm{J} \end{gathered}$$

Hence, the value of the work is -0.90 J.

Step 4(b): Calculation of the increase in the thermal energy of the system

The increase in thermal energy of the block –floor system occurs because of the friction between the block and the floor surface. Thus, the thermal energy of the system can be given:

$$\begin{gathered} KE\left(block\right)=PE\left(spring\right)+E_{th} \\ \frac{1}{2}m{v}^2=0.90+0.46 \\ =1.36 \\ {v}^2=\frac{1.36\times 2}{2.5} \\ =1.09 \\ v=1.0\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the thermal energy is 1.0 m/s

Step 5(c): Calculation of the speed of the block

Now, for getting the speed of the block, we use the conservation of energy principle,

$$\begin{gathered} \mathrm{KE}\left(\mathrm{block}\right)=\mathrm{PE}\left(\mathrm{spring}\right)+{\mathrm{E}}_{\mathrm{th}} \\ \frac{1}{2}{\mathrm{mv}}^2=0.90+0.46 \\ {\mathrm{v}}^2=\frac{1.36\times 2}{2.5} \\ =1.09 \\ \mathrm{v}=1.0\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the block is 1.0 m/s .