Question

4. (III) A 17,000-kg jet takes off from an aircraft carrier via a catapult (Fig. 6–39a). The gases thrust out from the jet’s engines exert a constant force of 130 kN on the jet; the force exerted on the jet by the catapult is plotted in Fig. 6–39b. Determine the work done on the jet: (a) by the gases expelled by its engines during launch of the jet; and (b) by the catapult during launch of the jet.

FIGURE 6–39 Problem 14

Short answer

(a) The work done on the jet by the gases expelled by its engines during the launch of the jet is \(1.10 \times {10^7}\;{\rm{J}}\).

(b) The work done by the catapult during the launch of the jet is \(4.95 \times {10^7}\;{\rm{J}}\).

Step-by-step solution

Calculation of work done by a constant force

The work done on an object by an applied constant force is equal to the object's displacement multiplied by the force component in the object's displacement direction.

Given information

Given data:

The constant force applied by the gases is\(F = 130\;{\rm{kN}}\).

The displacement is \(x = 85\;{\rm{m}}\).

Calculate the work done by the gases

(a)

The work done on the jet by the gases expelled by its engines during the launch of the jet can be calculated as:

\(\begin{aligned}{W_1} &= Fx\cos \left( {0^\circ } \right)\\{W_1} &= \left( {{\rm{130}}\;{\rm{kN}}} \right)\left( {\frac{{{\rm{1000}}\;{\rm{N}}}}{{{\rm{1}}\;{\rm{kN}}}}} \right)\left( {85\;{\rm{m}}} \right)\cos \left( {0^\circ } \right)\\{W_1} &= 1.10 \times {10^7}\;{\rm{J}}\end{aligned}\)

Thus, the work done on the jet by the gases expelled by its engines during the launch of the jet is \(1.10 \times {10^7}\;{\rm{J}}\).

Calculate the work done by the catapult

(b)

The work done by the catapult is equal to the area under the curve of forces versus displacement. Thus, the work done can be calculated as:

\(\begin{aligned}{W_2} &= \left\{ \begin{aligned}{l}\left( {{\rm{Area}}\;{\rm{of}}\;{\rm{triangle}}\;{\rm{of}}\;{\rm{base}}\;{\rm{85}}\;{\rm{m}}\;{\rm{and}}\;{\rm{height}}\;\left( {{\rm{1100}}\;{\rm{kN}} - {\rm{65}}\;{\rm{kN}}} \right)} \right)\\ + \left( {{\rm{Area}}\;{\rm{of}}\;{\rm{rectangular}}\;{\rm{of}}\;\left( {65\;{\rm{kN}} \times 85\;{\rm{m}}} \right)} \right)\end{aligned} \right\}\\{W_2} &= \frac{1}{2}\left( {85\;{\rm{m}}} \right)\left( {\left\{ {\left( {1100 - 65} \right)\;{\rm{kN}}} \right\} \times \left( {\frac{{{{10}^3}\;{\rm{N}}}}{{1\;{\rm{kN}}}}} \right)} \right) + \left( {\left( {65\;{\rm{kN}} \times \frac{{{{10}^3}\;{\rm{N}}}}{{1\;{\rm{kN}}}}} \right)\left( {85\;{\rm{m}}} \right)} \right)\\{W_2} &= 4.95 \times {10^7}\;{\rm{J}}\end{aligned}\)

Thus, the work done by the catapult is \(4.95 \times {10^7}\;{\rm{J}}\).