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Some of the funniest videos on the web involve motorists sliding uncontrollably on icy roads. Here let's compare the typical stopping distances for a car sliding to a stop from an initial speed of 10.0 m/s on a dry horizontal road, an icy horizontal road, and (everyone's favorite) an icy hill. (a) How far does the car take to slide to a stop on a horizontal road (Fig.6-4a) if the coefficient of kinetic friction is mu_(k)=0.60, which is typical of regular tires on dry pavement? Let's neglect any effect of the air on the car, assume that the wheels lock up and the tires slide, and extend an x axis in the car's direction of motion. |
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Answer» SOLUTION :The car accelerates (its speed decreases) because a horizontal frictional force acts againts the motion, in the negative direction of the x axis. (2) The frictional force is a kinetic frictional force with a magnitude of the given by Eq. 6-2 `(f_(k)=mu_(k)F_(N))`, in which `F_(N)` is the magnitude of the normal force on the car from the road. (3) We can relate the frictional force to the resulting ACCELERATION by writing Newton.s second law `(F_("net.x"=ma_(x)))` for motion along the road. Calculations: Figure 6-4b shows the free-body diagram for the car. The normal force is upward, the gravitational force is downward, and the frictional force is horizontal. Because the frictional force is the only force with an x component, Newton.s second law written for motion along the x axis becomes `-f_(k)=ma_(x)`  Figure 6-4 (a) A car sliding to the right and finally stopping. A free-body diagram for the car on (b) the same horizontal road and (c) a haill. Substituting `f_(k)=mu_(k)F_(N)` gives us `-mu_(k)F_(N)=ma_(x)`. Form Fig. 6-4b we SEE that the upward normal force balances the downward gravitational force, so in Eq. 6-9 let.s replace magnitude `F_(N)` with magnitude mg. Then we can cancel m (the stopping distance is thus independent of the car.s mass- the car can be heavy or light, it does not matter). Solving for `a_(x)` we find `a_(x)=-mu_(k)g`. Because this acceleration is constant, we can use the constant-acceleration equation of Table 2-1. The easiest choice for finding the sliding distance `x-x_(0)` is Eq. 2-16 `(v^(2)=v_(0)^(2)+2ax-x_(0))`, which gives us `x-x_(0)=(v^(2)-v_(0)^(2))/(2a_(x))` Substituting from Eq. 6-10 we then have `x-x_(0)=(v^(2)-v_(0)^(2))/(-2mu_(k)g)` Inserting the INITIAL speed `v_(0)=10.0` m/s, the final speed v=0, and the coefficient of kinetic friction `mu_(k)=0.60`, we find that the car.s stopping distance is `x-x_(0)=8.50m~~8.5m` (b) What is the stopping distance if the road is covered with ice with `mu_(k)=0.10`? Calculation: Our solution is perfectly fine through Eq. 6-12 but now we substitute this new `mu_(k)`, finding `x-x_(0)=51`m. Thus, a much longer clear path would be needed to avoid the car hitting something along the way. (c) Now let.s have the car sliding down an icy hill with an inclination of `theta=5.00^(@)` (a mild incline, nothing like the hills of San Francisco). The free-body diagram shown in Fig. 6-4c like the ramp in Sample Problem 5.06 except, to be consistent with Fig. 6-4b, the positive direction of the x axis is down the ramp. What now is the stopping distance? Calculations: Switching from Fig. 6-4b. to c involves two major changes. (1) Now a component of the gravitational force is along the tilted x axis, PULLING the car down the hill. From Sample Problem 5.06 and Fig 5-23, that down-the-hill component is mg `sintheta`, which is in the positive direction of the x axis in Fig. 6-4c. (2) The normal force (still perpendicular to the road) now balances only a component of the gravitational force, not the full force. From Sample Problem 5.04 (see Fig. 5-23i), we write that balance as `F_(N)=mgcostheta`. In spite of these changes, we still want to write Newton.s second law `(F_("net.x")=ma_(x))` for the motion along the (now tilted) x axis. We have `-f_(k)+mgsintheta=ma_(x)` `-mu_(k)F_(N)+mgsintheta=ma_(x)`, `-mu_(k)mgcostheta+mgsintheta=ma`, Solving for the acceleration and substituting the given data now give us `a_(x)=-mu_(k)gcostheta+gsintheta` `=-(0.10)(9.8m//s^(2))cos5.00^(@)+(9.8m//s^(2))sin5.00^(@)` `=-0.122m//s^(2)`. Substituting this result into Eq. 6-11 gives us the stopping distance down the hill: `x-x_(0)=409m~~400m` which is about 1/4 mile. Such icy hills separate people who can do this calculation (and thus know to stay home) from people who cannot (and thus end up in web videos). |
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