Chapter 7, Problem 93P

(a)

To determine

The initial kinetic energy of rock.

Explanation of Solution

Given:

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

The initial speed of rock piece is $40\text{m}/\text{s}$ .

The maximum height attained by the rock piece is $50\text{m}$ .

Formula used:

Write the expression for initial kinetic energy.

  $K_i=\frac{1}{2}m{v}_i^2$ ........(1)

Here, $m$ is mass of object, $K_i$ is the initial kinetic energy and $v_i$ is the initial velocity of object

Calculation:

Substitute $2.0\text{kg}$ for $m$ and $40\text{m}/\text{s}$ for $v_i$ in equation (1).

  $\begin{array}{c}{K}_i=\frac{1}{2}\left(2.0\text{kg}\right){\left(40\text{m}/\text{s}\right)}^2\\ =1600\text{J}\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ =1.6\text{kJ}\end{array}$

Conclusion:

Thus, the initial kinetic energy of rock pieceis $1.6\text{kJ}$ .

(b)

To determine

The increase in thermal energy due to air resistance.

Explanation of Solution

Given:

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

The initial speed of rock piece is $40\text{m}/\text{s}$ .

The maximum height attained by the rock piece is $50\text{m}$ .

Formula used:

Write the expression for change in potential energy.

  $\Delta U=mg\left(h_f-h_i\right)$

Here, $\Delta U$ is the change in potential energy, $m$ is mass of object, $g$ is gravitational acceleration, $h_i$ is the initial height of object and $h_f$ is the final height of object.

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 object and $v_f$ is the final velocity of object.

Total energy of object 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, $W_{\text{external}}$ is the work done by external force and $\Delta E_{\text{thermal}}$ is the thermal energy.

There is no external force acting on the block; so, the work done by external force is zero.

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

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

Substitute $0$ for $h_i$ , $0$ for $v_f$ and rearrange the above expression in terms of $\Delta E_{\text{thermal}}$ .

  $\Delta E_{\text{thermal}}=\frac{1}{2}m{v}_i^2-mg{h}_f$ ........(2)

Calculation:

Substitute $1.6\text{kJ}$ for $\frac{1}{2}m{v}_i^2$ , $2.0\text{kg}$ for $m$ , $9.81\text{m}/{\text{s}}^2$ for $g$ and $50\text{m}$ for $h_f$ in equation (2).

  $\begin{array}{c}\Delta E_{\text{thermal}}=1.6\text{kJ}-\left(2.0\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(50\text{m}\right)\\ =1.6\text{kJ}-\left(981\text{J}\right)\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ =0.619\text{kJ}\end{array}$

Conclusion:

Thus, the increase in thermal energy due to air resistance is $0.619\text{kJ}$ .

(c)

To determine

The final speed of the rock piece.

Explanation of Solution

Given:

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

The initial speed of rock piece is $40\text{m}/\text{s}$ .

The maximum height attained by the rock piece is $50\text{m}$ .

The increase in thermal energy during downward motion is $70\%$ to the thermal energy during upward motion.

Formula used:

Write the expression for change in potential energy.

  $\Delta U=mg\left(h_f-h_i\right)$

Here, $\Delta U$ is the change in potential energy, $m$ is mass of object, $g$ is gravitational acceleration, $h_i$ is the initial height of object and $h_f$ is the final height of object.

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 object and $v_f$ is the final velocity of object.

Total energy of object 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, $W_{\text{external}}$ is the work done by external force and $\Delta E_{\text{thermal}}$ is the thermal energy.

There is no external force acting on the block; so, the work done by external force is zero.

For rock-piece moving upward:

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

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

Substitute $0$ for $h_i$ , $0$ for $v_f$ and rearrange the above expression in terms of $\Delta E_{\text{thermal}}$ .

  $\Delta E_{\text{thermal}}=\frac{1}{2}m{v}_i^2-mg{h}_f$

For rock-piece moving downward:

Substitute $\frac{1}{2}m\left(v_f^2-v_i^2\right)$ for $\Delta K$ , $0$ for $W_{\text{external}}$ , $0.70\left(\Delta E_{\text{thermal}}\right)$ for $\Delta E_{\text{thermal}}$ and $mg\left(h_f-h_i\right)$ for $\Delta U$ in equation (1).

  $0=\frac{1}{2}m\left(v_f^2-v_i^2\right)+mg\left(h_f-h_i\right)+0.70\left(\Delta E_{\text{thermal}}\right)$

Substitute $0$ for $h_f$ , $0$ for $v_i$ and rearrange the above expression in terms of $v_f$ .

  $v_f=\sqrt{2\left(g{h}_i-\frac{0.70\left(\Delta E_{\text{thermal}}\right)}{m}\right)}$ ........(2)

Calculation:

Substitute $0.619\text{kJ}$ for $\Delta E_{\text{thermal}}$ , $2.0\text{kg}$ for $m$ , $9.81\text{m}/{\text{s}}^2$ for $g$ and $50\text{m}$ for $h_i$ in equation (2).

  $\begin{array}{c}{v}_f=\sqrt{2\left(\left(9.81\text{m}/{\text{s}}^2\right)\left(50\text{m}\right)-\frac{0.70\left(0.619\text{kJ}\right)}{2.0\text{kg}}\right)}\\ =23\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the final speed of the rock piece is $23\text{m}/\text{s}$ .