<p>This is problem on "Changes in Mechanical Energy for Nonconservative Forces." (Ch.8 #29 in Serway Jewett, Physics for Engineers)</p>
<p>This is what the question says verbatim:</p>
<p>The world's biggest locomotive is the MK5000C, a behemoth of mass 160 metric tons driven by the most powerful engine ever used for rail transportation, a Caterpillar diesel capable of 5000hp. Such a huge machine can provide a gain in efficiency, but its large mass presents challenges as well. The engineer finds that the locomotive handles differently from conventional units, notably in braking and climbing hills. Consider the locomotive pulling no train, but traveling at 27.0m/s on a level track while operating with output power 1000hp. It comes to a 5.00% grade (a slope that rises 5.00m for every 100m along the track). If the throttle is not advanced, so that the power level is held steady, to what value will the speed drop? Assume that friction forces do not depend on the speed.</p>
<p>Answer is 6.92m/s. I just wanna know how to get it.</p>
<p>I solved it and found the answer to be 7.04. I can steer you that close if you like. Sometimes the Serway answers can be a little wierd cuz of how they approach sig figs and trig functions.</p>
<p>Sure.
Remember the Work-Energy Theorum? Well, in this problem you are dealing with power. Since power is just work or energy divided by time, then there must be a way to use a formula just like the one for work and energy.
You already know the power going into the system (1000hp * 746Watts/HP), use the theorum to find the power consumed.
I used dimensional analysis to find how to include velocity into the work equation so that it would equal power</p>
<p>Remember, Force * Distance = Work<br>
Therefore, Force * Velocity (D/T) = Power. </p>
<p>So Power going in = power lost to frictional force + power used to climb the slope.
You can find the Frictional force using the 27 m/s speed.
Force * Velocity (27m/s) = Power in (746000 W)
Solve for friction force</p>
<p>The power lost to climbing the hill can be found using the expression mgsin(angle)v (What helped me find that was to draw the slope and train and drawn lines down for the normal force and the gravitational force. By drawing a line perpendicular to the normal force, a new triangle is found)</p>
<p>use hypotenuse = 100 and opposite = 5 to find an angle of 2.866</p>
<p>V is your variable. Solving for it I get 7.04 m/s (I graphed)</p>
<p>I would be interested to find out why Serway found something different. Anyway, this was fun and a was good review for me. I am self-studying AP Physics C from Serway's "Physics for Scientists and Engineers"</p>