/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 163 \( \mathrm{A}\) rocket sled has ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 163

\( \mathrm{A}\) rocket sled has mass of \(5000 \mathrm{kg}\). including \(1000 \mathrm{kg}\) of fuel. The motion resistance in the track on which the sled rides and that of the air total \(k U,\) where \(k\) is \(50 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}\) and \(U\) is the speed of the sled in \(\mathrm{m} / \mathrm{s}\). The exit speed of the exhaust gas relative to the rocket is \(1750 \mathrm{m} / \mathrm{s}\), and the exit pressure is atmospheric. The rocket burns fuel at the rate of \(50 \mathrm{kg} / \mathrm{s}\). (a) Plot the sled speed as a function of time. (b) Find the maximum speed. (c) What percentage increase in maximum speed would be obtained by reducing \(k\) by 10 percent?

Short Answer

Expert verified
a) speed of the sled as function of time is: \(U(t) = \frac {\mu V_e} {k}(1-e^{-kt/m})\); b) maximum speed is \(U_{max} = \frac {\mu V_e} {k}(1-e^{-kT/m})\); c) the percentage increase in maximum speed when reducing \(k\) by 10 percent is \(\frac{U'_{max} - U_{max}} {U_{max}} * 100\)

Step by step solution

01

Formulate the equation of motion for the sled

Sum up the forces acting on the sled. The force exerted by the rocket is \(F_{rocket} = \mu V_e\), where \(\mu\) is the rate at which fuel burns and \(V_e\) is the exit speed of the exhaust gas relative to the rocket. The resistive force is \(F_{resistance} = kU\), where \(k\) is the resistance coefficient and \(U\) is the speed of the sled. The equation of motion, from Newton's second law (Force = mass x acceleration), will be: \(Ma = \mu V_e - kU\), where \(M = m - \mu t \), \(m\) is the initial mass of the sled, and \(t\) is time.
02

solve the differential equation for sled speed as a function of time

The differential equation of motion, in terms of speed, U, and time, t, becomes a separable equation: \(M dU = (\mu V_e - kU) dt\). This equation can be integrated from \(t=0, U=0\) to \(t=T, U=U(T)\) to find an explicit expression of \(U\) in terms of \(t\), where \(T\) is the time when fuel runs out. The integration leads to the time-dependent speed of the sled: \(U(t) = \frac {\mu V_e} {k}(1-e^{-kt/m})\), where \(e\) is the base of natural logarithms.
03

Find the maximum speed

The maximum speed, \(U_{max}\), is reached when the rocket has used up all its fuel, at \(t=T\), where \(T = m / \mu\). Put these values into the equation from step 2 to find \(U_{max} = \frac {\mu V_e} {k}(1-e^{-kT/m})\).
04

Calculate the percentage increase in maximum speed when reducing the resistance coefficient by 10 percent

Replace \(k\) by 0.9\(k\) in the expression for the maximum speed to find \(U_{max}'\). The percentage increase is then \(\frac{U'_{max} - U_{max}} {U_{max}} * 100\)

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