Chapter 7, Problem 65P

(a)

To determine

The speed of the block at the bottom of the ramp.

Explanation of Solution

Given:

The mass of block is $2.0\text{kg}$ .

The initial height of block is $3.0\text{m}$ .

The final height of block is $0$ .

The distance traveled by block on horizontal surface is $9.0\text{m}$ .

Formula used:

Write the expression for change in potential energy.

  $\Delta U=mg\left(h_f-h_i\right)$

Here, $m$ is the mass of block, $h_i$ is the initial height of block and $h_f$ is the final height of block.

Write the expression for change in kinetic energy.

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

Here, $v_i$ is the initial velocity of block and $v_f$ is the final velocity of block.

Total energy of block is conserved at all points. Work done by external force is equal to the sum of change in gravitational potential energy, kinetic energy and thermal energy.

Write the expression of work done by external force.

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

Here, $\Delta U$ is the potential energy stored in the body, $\Delta K$ is the change in kinetic energy, $\Delta E_{\text{thermal}}$ is the thermal energy and $W_{\text{external}}$ is the work done by external force

Substitute $0$ for $W_{\text{external}}$ , $0$ for $\Delta E_{\text{thermal}}$ , $mg\left(h_f-h_i\right)$ for $\Delta U$ and $\frac{1}{2}m\left(v_f^2-v_i^2\right)$ for $\Delta K$ in equation (1).

  $0=mg\left(h_f-h_i\right)+\frac{1}{2}m\left(v_f^2-v_i^2\right)+0$

Rearrange the above expression in terms of $v_f$ .

  $v_f=\sqrt{2g\left(h_i-h_f\right)+v_i^2}$ ........(2)

Calculation:

Substitute $9.81\text{m}/{\text{s}}^2$ for $g$ , $3.0\text{m}$ for $h_i$ , $0$ for $h_f$ and $0$ for $v_i$ in equation (2).

  $\begin{array}{c}{v}_f=\sqrt{2\left(9.81\text{m}/{\text{s}}^2\right)\left(3.0\text{m}-0\right)+0}\\ =7.67\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the speed of the block at the bottom of the ramp is $7.67\text{m}/\text{s}$ .

(b)

To determine

The energy dissipated by the friction force.

Explanation of Solution

Given:

The mass of block is $2.0\text{kg}$ .

The initial height of block is $3.0\text{m}$ .

The final height of block is $0$ .

The distance traveled by block on horizontal surface is $9.0\text{m}$ .

Formula used:

Write the expression for change in potential energy.

  $\Delta U=mg\left(h_f-h_i\right)$

Here, $m$ is the mass of block, $h_i$ is the initial height of block and $h_f$ is the final height of block.

Write the expression for change in kinetic energy.

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

Here, $v_i$ is the initial velocity of block and $v_f$ is the final velocity of block.

Total energy of block is conserved at all points. Work done by external force is equal to the sum of change in gravitational potential energy, kinetic energy and frictional energy.

Write the expression of work done by external force.

  $W_{\text{external}}=\Delta U+\Delta K+W_{\text{friction}}$

Rearrange the above expression in terms of $W_{\text{friction}}$ .

  $W_{\text{friction}}=W_{\text{external}}-\Delta U-\Delta K$ ........(3)

Here, $\Delta U$ is the potential energy stored in the body, $\Delta K$ is the change in kinetic energy, $W_{\text{friction}}$ is the frictional energy and $W_{\text{external}}$ is the work done by external force.

Substitute $0$ for $W_{\text{external}}$ , $mg\left(h_f-h_i\right)$ for $\Delta U$ and $\frac{1}{2}m\left(v_f^2-v_i^2\right)$ for $\Delta K$ in equation (3).

  $W_{\text{friction}}=-\left[mg\left(h_f-h_i\right)+\frac{1}{2}m\left(v_f^2-v_i^2\right)\right]$ ........(4)

Calculation:

Substitute $2.0\text{kg}$ for $m$ , $9.81\text{m}/{\text{s}}^2$ for $g$ , $3.0\text{m}$ for $h_i$ , $0$ for $h_f$ , $0$ for $v_i$ and $0$ for $v_f$ in equation (4).

  $\begin{array}{c}{W}_{\text{friction}}=-\left[\left(2.0\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(0-3.0\text{m}\right)+\frac{1}{2}\left(2.0\text{kg}\right)\left(0-0\right)\right]\\ =58.96\text{J}\end{array}$

Conclusion:

Thus, the energy dissipated by the friction force is $58.96\text{J}$ .

(c)

To determine

The coefficient of kinetic friction between the block and the horizontal surface.

Explanation of Solution

Given:

The mass of block is $2.0\text{kg}$ .

The initial height of block is $3.0\text{m}$ .

The final height of block is $0$ .

The distance traveled by block on horizontal surface is $9.0\text{m}$ .

Formula used:

Write the expression for change in potential energy.

  $\Delta U=mg\left(h_f-h_i\right)$

Here, $m$ is the mass of block, $h_i$ is the initial height of block and $h_f$ is the final height of block.

Write the expression for change in kinetic energy.

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

Here, $v_i$ is the initial velocity of block and $v_f$ is the final velocity of block.

Total energy of block is conserved at all points. Work done by external force is equal to the sum of change in gravitational potential energy, kinetic energy and frictional energy.

Write the expression of work done by external force.

  $W_{\text{external}}=\Delta U+\Delta K+W_{\text{friction}}$

Rearrange the above expression in terms of $W_{\text{friction}}$ .

  $W_{\text{friction}}=W_{\text{external}}-\Delta U-\Delta K$ ........(3)

Here, $\Delta U$ is the potential energy stored in the body, $\Delta K$ is the change in kinetic energy, $W_{\text{friction}}$ is the frictional energy and $W_{\text{external}}$ is the work done by external force.

Substitute $0$ for $W_{\text{external}}$ , $mg\left(h_f-h_i\right)$ for $\Delta U$ and $\frac{1}{2}m\left(v_f^2-v_i^2\right)$ for $\Delta K$ in equation (3).

  $W_{\text{friction}}=-\left[mg\left(h_f-h_i\right)+\frac{1}{2}m\left(v_f^2-v_i^2\right)\right]$ ........(4)

Write the expression for friction force.

  $f=\mu mg$

Here, $f$ is the friction force and $\mu$ is the coefficient of kinetic friction.

Rearrange the above expression in terms of $\mu$ .

  $\mu =\frac{f}{mg}$ ........(5)

Write the expression for frictional energy.

  $W_{\text{friction}}=f\Delta s$

Here, $W_{\text{friction}}$ is the thermal energy produced and $\Delta s$ is the displacement covered by block.

Rearrange the above expression in terms of $f$ .

  $f=\frac{W_{\text{friction}}}{\Delta s}$

Substitute $W_{\text{friction}}/\Delta s$ for $f$ in equation (5).

  $\mu =\frac{W_{\text{friction}}}{mg\Delta s}$ ........(6)

Calculation:

Substitute $2.0\text{kg}$ for $m$ , $9.81\text{m}/{\text{s}}^2$ for $g$ , $3.0\text{m}$ for $h_i$ , $0$ for $h_f$ , $0$ for $v_i$ and $0$ for $v_f$ in equation (4).

  $\begin{array}{c}{W}_{\text{friction}}=-\left[\left(2.0\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(0-3.0\text{m}\right)+\frac{1}{2}\left(2.0\text{kg}\right)\left(0-0\right)\right]\\ =58.96\text{J}\end{array}$

Substitute $2.0\text{kg}$ for $m$ , $9.81\text{m}/{\text{s}}^2$ for $g$ , $9.0\text{m}$ for $\Delta s$ and $58.96\text{J}$ for $W_{\text{friction}}$ in equation (6).

  $\begin{array}{c}\mu =\frac{58.96\text{J}}{\left(2.0\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(9.0\text{m}\right)}\\ =0.33\end{array}$

Conclusion:

Thus, the coefficient of kinetic friction between the block and the horizontal surface is $0.33$ .