Chapter 5, Problem 65P

(a)

To determine

To show coefficient of kinetic friction ${\mu }_k=\frac{2\Delta x}{{\left(\Delta t\right)}^{2}g}$ .

Explanation of Solution

Given:

The time measure is $\Delta t$ .

Total displacement of the block is $\Delta x$ .

Formula used:

Draw free body diagram of block.

  

Write expression for net force in the x -direction.

  ${\displaystyle \sum F_x}=0$

Substitute $-f_k-m{a}_x$ for ${\displaystyle \sum F_x}$ in above expression.

  $-f_k-m{a}_x=0$

Here, $f_k$ is the friction for due to kinetic friction, $m$ is the mass of the block and $a_x$ is the acceleration of the block.

Rearrange above expression for $f_k$ .

  $f_k=-m{a}_x$

Substitute ${\mu }_{k}mg$ for $f_k$ in above expression and solve for $a_x$ .

  $\begin{array}{c}{\mu }_{k}mg=-m{a}_x\\ a_x=-{\mu }_{k}g\end{array}$

Here, ${\mu }_k$ is the coefficient of the kinetic friction and $g$ is the acceleration due to gravity.

Calculation:

Write expression for displacement of the block.

  $\Delta x=v_{av}\Delta t$

Here, $v_{av}$ is the average velocity of the block and $\Delta t$ is the time taken by the block.

Substitute $\left(\frac{v_1+v_2}{2}\right)$ for $v_{av}$ in above expression.

  $\Delta x=\left(\frac{v_1+v_2}{2}\right)\Delta t$

Here, $v_1$ is the initial speed of the block and $v_2$ is the final speed of the block.

Substitute $0$ for $v_2$ in above expression and solve for $v_1\Delta t$ .

  $\begin{array}{c}\Delta x=\left(\frac{v_1+0}{2}\right)\Delta t\\ v_1\Delta t=2\Delta x\end{array}$

Write expression for the displacement of the block.

  $\Delta x=v_1\Delta t+\frac{1}{2}{a}_x{\left(\Delta t\right)}^2$

Substitute $2\Delta x$ for $v_1\Delta t$ in above expression.

  $\begin{array}{c}\Delta x=2\Delta x+\frac{1}{2}{a}_x{\left(\Delta t\right)}^2\\ \Delta x=-\frac{1}{2}{a}_x{\left(\Delta t\right)}^2\end{array}$

Substitute $-{\mu }_{k}g$ for $a_x$ in above expression and solve for ${\mu }_k$ .

  $\begin{array}{c}\Delta x=-\frac{1}{2}\left(-{\mu }_{k}g\right){\left(\Delta t\right)}^2\\ {\mu }_k=\frac{2\Delta x}{g{\left(\Delta t\right)}^2}\end{array}$

Conclusion:

Thus, the coefficient of kinetic friction is ${\mu }_k=\frac{2\Delta x}{{\left(\Delta t\right)}^{2}g}$ .

(b)

To determine

The value of coefficient of kinetic friction.

Explanation of Solution

Given:

The block slides $1.37\text{m}$ .

The time taken is $0.97\text{s}$ .

Formula used:

Draw free body diagram of block.

  

Write expression for net force in the x -direction.

  ${\displaystyle \sum F_x}=0$

Substitute $-f_k-m{a}_x$ for ${\displaystyle \sum F_x}$ in above expression.

  $-f_k-m{a}_x=0$

Rearrange above expression for $f_k$ .

  $f_k=-m{a}_x$

Substitute ${\mu }_{k}mg$ for $f_k$ in above expression and solve for $a_x$ .

  $\begin{array}{c}{\mu }_{k}mg=-m{a}_x\\ a_x=-{\mu }_{k}g\end{array}$

Write expression for displacement of the block.

  $\Delta x=v_{av}\Delta t$

Substitute $\left(\frac{v_1+v_2}{2}\right)$ for $v_{av}$ in above expression.

  $\Delta x=\left(\frac{v_1+v_2}{2}\right)\Delta t$

Substitute $0$ for $v_2$ in above expression and solve for $v_1\Delta t$ .

  $\begin{array}{c}\Delta x=\left(\frac{v_1+0}{2}\right)\Delta t\\ v_1\Delta t=2\Delta x\end{array}$

Write expression for the displacement of the block.

  $\Delta x=v_1\Delta t+\frac{1}{2}{a}_x{\left(\Delta t\right)}^2$

Substitute $2\Delta x$ for $v_1\Delta t$ in above expression.

  $\begin{array}{c}\Delta x=2\Delta x+\frac{1}{2}{a}_x{\left(\Delta t\right)}^2\\ \Delta x=-\frac{1}{2}{a}_x{\left(\Delta t\right)}^2\end{array}$

Substitute $-{\mu }_{k}g$ for $a_x$ in above expression and solve for ${\mu }_k$ .

  ${\mu }_k=\frac{2\Delta x}{g{\left(\Delta t\right)}^2}$.......(1)

Calculation:

Substitute $1.37\text{m}$ for $\Delta x$ , $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.97\text{s}$ for $\Delta t$ in equation (1).

  $\begin{array}{c}{\mu }_k=\frac{2\left(1.37\text{m}\right)}{\left(9.81\text{m}/{\text{s}}^{\text{2}}\right){\left(0.97\text{s}\right)}^2}\\ =0.30\end{array}$

Conclusion:

Thus, the value of the kinetic friction is $0.30$ .

(c)

To determine

The initial speed of the block

Explanation of Solution

Given:

Formula used:

Draw free body diagram of block.

  

Write expression for net force in the x -direction.

  ${\displaystyle \sum F_x}=0$

Substitute $-f_k-m{a}_x$ for ${\displaystyle \sum F_x}$ in above expression.

  $-f_k-m{a}_x=0$

Rearrange above expression for $f_k$ .

  $f_k=-m{a}_x$

Substitute ${\mu }_{k}mg$ for $f_k$ in above expression and solve for $a_x$ .

  $\begin{array}{c}{\mu }_{k}mg=-m{a}_x\\ a_x=-{\mu }_{k}g\end{array}$

Write expression for displacement of the block.

  $\Delta x=v_{av}\Delta t$

Substitute $\left(\frac{v_1+v_2}{2}\right)$ for $v_{av}$ in above expression.

  $\Delta x=\left(\frac{v_1+v_2}{2}\right)\Delta t$

Substitute $0$ for $v_2$ in above expression and solve for $v_1$ .

  $v_1=\frac{2\Delta x}{\Delta t}$.......(1)

Calculation:

Substitute $1.37\text{m}$ for $\Delta x$ , and $0.97\text{s}$ for $\Delta t$ in equation (1).

  $\begin{array}{c}{v}_1=\frac{2\left(1.37\text{m}\right)}{0.97\text{s}}\\ =2.8\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the initial speed of the block is $2.8\text{m}/\text{s}$ .