Chapter 7, Problem 89P

(a)

To determine

The expression for the distance $d_1$ the box slides.

Explanation of Solution

Introduction:

To derive the expression for distance the box slides, the work-energy theorem will be used. Work-energy theorem states that work done by any external force on a body is equal to the change in total energy of that body.

Write the expression for friction force.

  $f=\mu mg$

Here, $m$ is mass of block, $g$ is gravitational acceleration, $\mu$ is the coefficient of friction and $f$ is the friction force.

Write the expression for thermal energy.

  $\Delta E_{\text{thermal}}=f\Delta x$

Here, $\Delta E_{\text{thermal}}$ is the thermal energy produced and $\Delta x$ is the displacement covered by block.

Substitute $\mu mg$ for $f$ in above expression.

  $\Delta E_{\text{thermal}}=\mu mg\Delta x$

Write the expression for change in kinetic energy.

  $\Delta K=\frac{1}{2}m\left(v_f^2-v_i^2\right)$

Here, $\Delta K$ is the change in kinetic energy, $v_i$ is the initial velocity of block and $v_f$ is the final velocity of block.

Write the expression for change in gravitational potential energy.

  $\Delta U_g=mg\left(h_f-h_i\right)$

Here, $\Delta U_g$ is change in gravitational potential energy, $h_i$ is the initial height of block and $h_f$ is the final height of block.

Write the expression for change in spring potential energy.

  $\Delta U_s=\frac{1}{2}k\left(x_f^2-x_i^2\right)$

Here, $\Delta U_s$ is change in spring potential energy, $k$ is the force constant of spring, $x_i$ is the initial compression ofspring and $x_f$ is the final compression of spring.

Write the expression of work done by external force.

  $W_{\text{external}}=\Delta U_g+\Delta U_s+\Delta K+\Delta E_{\text{thermal}}$ ........(1)

Here, $W_{\text{external}}$ is the work done by external force.

Substitute $0$ for $W_{\text{external}}$ , $0$ for $\Delta U_g$ , $0$ for $\Delta K$ , $\frac{1}{2}k\left(x_f^2-x_i^2\right)$ for $\Delta U_s$ and $\mu mg\Delta x$ for $\Delta E_{\text{thermal}}$ in equation (1).

  $0=0+\frac{1}{2}k\left(x_f^2-x_i^2\right)+0+\mu mg\Delta x$

Substitute $d_1-d_0$ for $x_f$ , $d_0$ for $x_i$ and $d_1$ for $\Delta x$ in above expression and rearrange in terms of $d_1$ .

  $d_1=2d_0-\frac{2\mu mg}{k}$

Conclusion:

Thus, the expression for the distance $d_1$ the box slides is $d_1=2d_0-\frac{2\mu mg}{k}$ .

(b)

To determine

The expression of speed for the box slides the distance $d_0$ .

Explanation of Solution

Introduction:

To derive the expression for distance the box slides, the work-energy theorem will be used. Work-energy theorem states that work done by any external force on a body is equal to the change in total energy of that body.

Write the expression for friction force.

  $f=\mu mg$

Here, $m$ is mass of block, $g$ is gravitational acceleration, $\mu$ is the coefficient of friction and $f$ is the friction force.

Write the expression for thermal energy.

  $\Delta E_{\text{thermal}}=f\Delta x$

Here, $\Delta E_{\text{thermal}}$ is the thermal energy produced and $\Delta x$ is the displacement covered by block.

Substitute $\mu mg$ for $f$ in above expression.

  $\Delta E_{\text{thermal}}=\mu mg\Delta x$

Write the expression for change in kinetic energy.

  $\Delta K=\frac{1}{2}m\left(v_f^2-v_i^2\right)$

Here, $\Delta K$ is the change in kinetic energy, $v_i$ is the initial velocity of block and $v_f$ is the final velocity of block.

Write the expression for change in gravitational potential energy.

  $\Delta U_g=mg\left(h_f-h_i\right)$

Here, $\Delta U_g$ is change in gravitational potential energy, $h_i$ is the initial height of block and $h_f$ is the final height of block.

Write the expression for change in spring potential energy.

  $\Delta U_s=\frac{1}{2}k\left(x_f^2-x_i^2\right)$

Here, $\Delta U_s$ is change in spring potential energy, $k$ is the force constant of spring, $x_i$ is the initial compression ofspring and $x_f$ is the final compression of spring.

Write the expression of work done by external force.

  $W_{\text{external}}=\Delta U_g+\Delta U_s+\Delta K+\Delta E_{\text{thermal}}$ ........(1)

Here, $W_{\text{external}}$ is the work done by external force.

Substitute $0$ for $W_{\text{external}}$ , $0$ for $\Delta U_g$ , $\frac{1}{2}m\left(v_f^2-v_i^2\right)$ for $\Delta K$ , $\frac{1}{2}k\left(x_f^2-x_i^2\right)$ for $\Delta U_s$ and $\mu mg\Delta x$ for $\Delta E_{\text{thermal}}$ in equation (1).

  $0=0+\frac{1}{2}k\left(x_f^2-x_i^2\right)+\frac{1}{2}m\left(v_f^2-v_i^2\right)+\mu mg\Delta x$

Substitute $0$ for $x_f$ , $d_0$ for $x_i$ , $0$ for $v_i$ , $v_0$ for $v_f$ and $d_0$ for $\Delta x$ in above expression and rearrange in terms of $v_0$ .

  $v_0=\sqrt{\frac{k}{m}{d}_0^2-2\mu g{d}_0}$

Conclusion:

Thus, the expression of speed for the box slides the distance $d_0$ is $v_0=\sqrt{\frac{k}{m}{d}_0^2-2\mu g{d}_0}$ .

(a)

To determine

The value of coefficient of friction for $d_1=d_0$ .

Explanation of Solution

Introduction:

To derive the expression for distance the box slides, the work-energy theorem will be used. Work-energy theorem states that work done by any external force on a body is equal to the change in total energy of that body.

Write the expression for friction force.

  $f=\mu mg$

Here, $m$ is mass of block, $g$ is gravitational acceleration, $\mu$ is the coefficient of friction and $f$ is the friction force.

Write the expression for thermal energy.

  $\Delta E_{\text{thermal}}=f\Delta x$

Here, $\Delta E_{\text{thermal}}$ is the thermal energy produced and $\Delta x$ is the displacement covered by block.

Substitute $\mu mg$ for $f$ in above expression.

  $\Delta E_{\text{thermal}}=\mu mg\Delta x$

Write the expression for change in kinetic energy.

  $\Delta K=\frac{1}{2}m\left(v_f^2-v_i^2\right)$

Here, $\Delta K$ is the change in kinetic energy, $v_i$ is the initial velocity of block and $v_f$ is the final velocity of block.

Write the expression for change in gravitational potential energy.

  $\Delta U_g=mg\left(h_f-h_i\right)$

Here, $\Delta U_g$ is change in gravitational potential energy, $h_i$ is the initial height of block and $h_f$ is the final height of block.

Write the expression for change in spring potential energy.

  $\Delta U_s=\frac{1}{2}k\left(x_f^2-x_i^2\right)$

Here, $\Delta U_s$ is change in spring potential energy, $k$ is the force constant of spring, $x_i$ is the initial compression ofspring and $x_f$ is the final compression of spring.

Write the expression of work done by external force.

  $W_{\text{external}}=\Delta U_g+\Delta U_s+\Delta K+\Delta E_{\text{thermal}}$ ........(1)

Here, $W_{\text{external}}$ is the work done by external force.

Substitute $0$ for $W_{\text{external}}$ , $0$ for $\Delta U_g$ , $0$ for $\Delta K$ , $\frac{1}{2}k\left(x_f^2-x_i^2\right)$ for $\Delta U_s$ and $\mu mg\Delta x$ for $\Delta E_{\text{thermal}}$ in equation (1).

  $0=0+\frac{1}{2}k\left(x_f^2-x_i^2\right)+0+\mu mg\Delta x$

Substitute $0$ for $x_f$ , $d_0$ for $x_i$ and $d_0$ for $\Delta x$ in above expression and rearrange in terms of $\mu$ .

  $\mu =\frac{k{d}_0}{2mg}$

Conclusion:

Thus, the value of coefficient of friction for $d_1=d_0$ is $\mu =\frac{k{d}_0}{2mg}$ .