91Ó°ÊÓ

We have given that:

The mass of the rocket (\(m\)) = 1500 kg.

Initial speed of the rocket (\({v_1}\)) = 0 m/s.

Final speed of the rocket (\({v_2}\)) = 50.0 m/s.

The angle between the ramp and the horizontal \(\theta \)= \(53^\circ \).

The thrust of the engine \(F\)= 2000 \(N\).

Friction force \({f_k}\) = 500 \(N\).

If we take \(h = 0\) at the bottom of the ramp, then the initial height is given by,

\({h_1} = d\,\sin \theta \), where \(d\) is the length of the ramp from the base to the starting point.

The final height is \({h_2} = 0\).

We know the work-energy theorem given by,

\({K_1} + {U_1} + W = {K_2} + {U_2}\,\,\, \cdots \cdots \left( 1 \right)\) .

Where the kinetic energy is given by,

\(K = \frac{1}{2}m{v^2}\,\, \cdots \cdots \left( 2 \right)\)

And the gravitational potential energy is given by,

\(U = mgh\,\,\, \cdots \cdots \left( 3 \right)\).

We know that the other two forces are the friction force which acts in the opposite direction to the direction of motion and the engine’s thrust which acts in the same direction of motion.

Therefore, the total work done is given by,

\(W = \left( {F - {f_k}} \right) \cdot d\,\,\, \cdots \cdots \left( 4 \right)\)

Now, substituting the values in \(\left( 4 \right)\) from step 2, we get,

\(\begin{aligned}{c}W = \left( {2000\;N - 500\;N} \right) \cdot d\\W = 1500\;N \cdot d\end{aligned}\)

Let us calculate energy quantities.

Since the rocket starts from the rest, we get,

\({K_1} = 0\;J\).

From \(\left( 3 \right)\), substituting the values of \(m,\,\;and\;{h_1}\), we get,

\(\begin{aligned}{c}{U_1} = 1500\;N \times 9.8\;{\mathord{\left/{\vphantom {m {{s^2}}}} \right. \\ {{s^2}}} \times \left( {d\sin 53^\circ } \right)\\{U_1} = \left( {11740\;{{N \cdot m} \mathord{\left/ {\vphantom {{N \cdot m} {{s^2}}}} \right. \\ {{s^2}}}} \right) \times d\end{aligned}\)

Now, substituting the values of \(m,\,\,and\;{v_2}\) in \(\left( 2 \right)\), we get,

\(\begin{aligned}{c}{K_2} = \frac{1}{2} \times 1500\;N \times {\left( {50\;{m \mathord{\left/ {\vphantom {m s}} \right. \\ s}} \right)^2}\\{K_2} = 1.86 \times {10^6}\;J\end{aligned}\)

Since the rocket ends at \(h = 0\), we get \({U_2} = 0\).

Finally, substituting all these values in \(\left( 1 \right)\), we get,

\(\begin{aligned}{l}0 + \left( {11740\;{{N \cdot m} \mathord{\left/{\vphantom {{N \cdot m} {{s^2}}}} \right.

\\{{s^2}}}}\right)\timesd+\left({1500\;{{N\cdotm}\mathord{\left/{\vphantom {{N \cdot m} {{s^2}}}} \right.

\\{{s^2}}}} \right) \times d = 1.86 \times {10^6}\;J + 0\;J\\13240 \times d = 1.86 \times {10^6}\;J\\d = \frac{{1.86 \times {{10}^6}\;J}}{{13240\;{{N \cdot m} \mathord{\left/ {\vphantom {{N \cdot m} {{s^2}}}} \right.

\\{{s^2}}}}}\\d = 141.6\;m\end{aligned}\)