Question

Consider the block–spring collision discussed in Example 8.8. (a) For the situation in part (B), in which the surface exerts a friction force on the block, show that the block never arrives back at x = 0. (b) What is the maximum value of the coefficient of friction that would allow the block to return to x = 0?

Short answer

The maximum value of the coefficient of friction that would allow the block to return to$\mathrm{x}=0$ is ${\mathrm{\mu }}_{\mathrm{k}}=\mathrm{0.342.}$

Step-by-step solution

Step 1:

In the presence of non-conservative forces, the total change in the mechanical energy of a system should be equal to the work done by the non-conservative forces.

$\sum {\mathbf{W}}_{\mathrm{other}\mathrm{forces}}\mathbf{=}\mathbf{W}\mathbf{=}\mathbf{\Delta K}\mathbf{+}\mathbf{\Delta U}\mathbf{+}{\mathbf{\Delta E}}_{\mathrm{int}}..........(\mathbf{8}\mathbf{.}\mathbf{17})$

Here:

$\mathbf{W}\mathbf{=}$The work by body

$\mathbf{\Delta U}\mathbf{=}$Change in Potential energy

$\mathbf{\Delta K}\mathbf{=}$Change in kinetic energy

${\mathbf{\Delta E}}_{\mathrm{int}}$ Change in internal energy

The potential energy of the system is:

$\mathbf{U}\mathbf{=}\frac{1}{2}{\mathbf{kx}}^2$

The kinetic energy of the system is:

$\mathbf{K}\mathbf{=}\frac{1}{2}{\mathbf{mv}}^2$

The definition of power is role="math" localid="1663649866131" $\mathbf{P}\mathbf{=}\mathbf{Fv}\mathbf{,}$ where:

$\mathbf{P}\mathbf{=}$Power of object

$\mathbf{F}\mathbf{=}$Force exerted on object

$\mathbf{v}\mathbf{=}$Speed of object

Step 2:

Part (a):

At the end of the process analyzed in Example 8.8, we begin with a $0.800\mathrm{kg}$ block at rest on the end of a spring with stiffness constant$50.0\mathrm{N}/\mathrm{m}$, compressed$0.0924\mathrm{m}$. The energy in the spring is:

$\begin{array}{c}\mathrm{U}=\frac{1}{2}{\mathrm{kx}}^2\\ =\frac{1}{2}(50.0\text{N/m}){(0.0924\text{m})}^2\\ =0.214\text{J.}\end{array}$

To push the block back to the unstressed spring position would require work against friction of magnitude

$\begin{array}{c}\mathrm{W}=\mathrm{Fm}\\ =(3.92\text{N})(0.0924\text{m})\\ =0.214\text{J.}\end{array}$

Because $0.214\mathrm{J}$ is less than $0.362\mathrm{J}$, the spring cannot push the object back to $\mathrm{x}=0$.

Part (b):

The block approaches the spring with energy.

$\begin{array}{c}\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}(0.800\text{kg}){(1.20\text{m/s})}^2\\ =0.576\text{J.}\end{array}$

As the block has to do work against the friction, so some of the initial kinetic energy is transformed to internal energy. At the maximum compression, the spring will only have half of the energy. So:

$\begin{array}{c}\frac{\mathrm{U}}{2}=\frac{1}{2}{\mathrm{kx}}^2\\ \frac{0.576\text{J}}{2}=\frac{1}{2}(50.0\text{N/m}){\mathrm{x}}^2\\ \mathrm{x}=0.107\text{m.}\end{array}$

Applying the law of conservation of energy in the compression process, we get:

$\begin{array}{c}0.576\text{J}+{\mathrm{\mu }}_{\mathrm{k}}7.84\text{N}(0.0107\text{m})\cos {180}^{\mathrm{o}}=0.288\text{J}\\ {\mathrm{\mu }}_{\mathrm{k}}=0.288\text{J}/0.841\text{J}\\ =\mathrm{0.342.}\end{array}$

The maximum coefficient of friction for the block to return to is ${\mathrm{\mu }}_{\mathrm{k}}=0.342$.