91影视

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 data:

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

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

(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 \(1.10 \times {10^7}\;{\rm{J}}\).

(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 \(4.95 \times {10^7}\;{\rm{J}}\).