Chapter 6, Problem 103QAP

To determine

(a)

The speed of the material as it leaves the surface.

Answer to Problem 103QAP

The speed of the material as it leaves the surface is $23.56\text{m/sec}$.

Explanation of Solution

Given:

  $\begin{array}{l}{g}_{mars}=3.7{\text{m/sec}}^{\text{2}}\\ h=75.0\text{m}\end{array}$

Concept Used:

Law of conservation of mechanical energy.

Calculation:

From the law of conservation of mechanical energy, we can calculate the speed of the material as it leaves the surface. The gravitational potential energy of the material at the surface is to be zero.

As the gas is travelling just below the surface of the surface of the planet, we can ignore the change in gravitational potential energy

According to law of conservation of energy:

  $\begin{array}{l}{U}_i+K_i=U_f+K_f\\ 0+\frac{1}{2}m{v}_i^2=m{g}_{mars}h+0\\ v_i=\sqrt{2g_{mars}h}\end{array}$

On replacing the values, we get

  $\begin{array}{l}{g}_{mars}=3.7{\text{m/sec}}^{\text{2}}\\ h=75.0\text{m}\\ v_i=\sqrt{2\times 3.7\times 75.0}\\ v_i=23.56\text{m/sec}\end{array}$

Conclusion:

The speed of the material as it leaves the surface is $23.56\text{m/sec}$.

To determine

(b)

The energy per kilogram of material is lost due to nonconservative forces.

Answer to Problem 103QAP

The energy per kilogram of material is lost due to nonconservative forces is $-7.0\times {10}^2\text{J/kg}$.

Explanation of Solution

Given:

  $\begin{array}{l}{g}_{mars}=3.7{\text{m/sec}}^{\text{2}}\\ h=75.0\text{m}\\ v_i=160\text{km/hr}\end{array}$

Concept Used:

Law of nonconservative of work.

Calculation:

From the law of conservation of mechanical energy, we can calculate the speed of the material as it leaves the surface. The gravitational potential energy of the material at the surface is to be zero.

As the gas is travelling just below the surface of the surface of the planet, we can ignore the change in gravitational potential energy.

As the gas leaves the jets it is just travelling underground at the speed of $160\text{km/hr}$

So, $v_i=160\text{km/hr}$

Converting the units:

  $\begin{array}{l}{v}_i=160\text{km/hr}\times \frac{{10}^3\text{m}}{1\text{km}}\times \frac{1\text{h}}{3600\sec }\\ v_i=44.4\text{m/sec}\end{array}$

According to law of conservation of energy:

  $\begin{array}{l}{U}_i+K_i=U_f+K_f\\ m{g}_{mars}h+0=0+\frac{1}{2}m{v}_f^2\\ v_f=\sqrt{2g_{mars}h}\end{array}$

On replacing the values, we get

  $\begin{array}{l}{g}_{mars}=3.7{\text{m/sec}}^{\text{2}}\\ h=75.0\text{m}\\ v_f=\sqrt{2\times 3.7\times 75.0}\\ v_f=23.56\text{m/sec}\end{array}$

According to law of nonconservative of work:

  $\begin{array}{l}{W}_{nonconservative}=\Delta K+\Delta U\\ W_{nonconservative}=\frac{1}{2}m(v_f^2-v_i^2)+0\\ E=\frac{W}{m}\\ E=\frac{W_{nonconservative}}{m}=\frac{1}{2}(v_f^2-v_i^2)\end{array}$

On replacing the values, we get

  $\begin{array}{l}{v}_i=44.4\text{m/sec}\\ v_f=23.56\text{m/sec}\\ E=\frac{1}{2}({23.56}^2-{44.4}^2)\\ E=-7.0\times {10}^2\text{J/kg}\end{array}$

Conclusion:

The energy per kilogram of material is lost due to nonconservative forces is $-7.0\times {10}^2\text{J/kg}$.