A student is racing along in a sports car when suddenly, to avoid an impending collision, the driver slams on the brakes and skids along a straight line 200 ft to a stop in a 45 mph zone. Moments earlier, a policeman had checked the vehicle’s speed on radar. Assume v = 0.6, ignore air resistance,

Question

A student is racing along in a sports car when suddenly, to avoid an impending collision, the driver slams on the brakes and skids along a straight line 200 ft to a stop in a 45 mph zone. Moments earlier, a policeman had checked the vehicle’s speed on radar. Assume v = 0.6, ignore air resistance, and determine if the officer might give the student a ticket for exceeding the limit. Show for this example that the work-energy equation is the first integral of the equation of motion.

Accepted Answer

The assigned speed data in Fig. 7.6 show that the change in the kinetic energy of the car and its driver of total mass m is given by [latex]\Delta K=-\frac{1}{2} m v^2[/latex], and we want to find v. Because the path and the position varying nature of the forces acting on the system are known, we consider the work–energy method.

The forces that act on the car are shown in the free body diagram of Fig. 7.6. Both N and W do no work in the motion, and their magnitudes are equal: N = W. The work done by the nonconservative Coulomb frictional force [latex]\mathbf{f}_d=-v N \mathbf{i}[/latex] in the plane motion along a straight line is determined by (7.30) in which d = l, namely, [latex]\mathscr{W}_f=-v W \ell[/latex]. The work–energy principle (7.36) thus yields the result [latex]-\frac{1}{2} m v^2=-v W \ell[/latex], and with W = mg, the initial speed is determined by

[latex]\mathscr{W}_f=-vNd=-f_dd.[/latex] (7.30)

[latex]\mathscr{W}=\Delta K[/latex] (7.36)

[latex]v=\sqrt{2 v g \ell}[/latex], (7.41a)

which is independent of the weight of the vehicle and its passenger.

For the given data, [latex]v=[2(.6)(32.2)(200)]^{1 / 2}=87.91 \mathrm{ft} / \mathrm{sec}[/latex], or very nearly 60 mph. In consequence, the student could receive a citation for speeding.

The first integral of the equation of motion [latex]m\ddot{s} =d(\frac{1}{2 }m\dot{s}^{2} )/ds=-vmg[/latex] yields the general form of work–energy equation:

[latex]\frac{1}{2}m\dot{s}^{2}- \frac{1}{2}mv^{2}=-vmgs. [/latex] (7.41b)

When s =ℓ , [latex]\dot{s} =0[/latex], we obtain (7.41a). Conversely, differentiation of the work–energy equation (7.41b) with respect to either s or t yields the equivalent equation of motion .

Q. 7.6

Chapter 7

Q. 7.6

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