7.25 | How high a hill can a car coast up (engine disengaged) if work done by friction is negligible
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- Опубліковано 25 сер 2024
- (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h? (b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction? (c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?
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Few Points: 1) Answer for (a) should be 47.6 m if you don't round your calculation until the end. 2) Give the absolute value of the answer in the video for (b) since it is asking for thermal energy and should be independent of sign. 3) The answer should be negative for (c). Technically, the force that is plugged into and spit out of W=|F|d*cos(theta) is the absolute value of the force; otherwise you would be DOUBLE COUNTING the direction of the vectors, since cosine is also involved. Therefore, when you arrive at your answer, it will be without a sign and you must then consider the direction of the force relative to the coordinate system you were using for the problem. Since it is assumed that left, or down the slope, is negative, then the force of friction should also be negative. 4) Don't sweat the small stuff. 5) We all make mistakes. 6) It's what you do about those mistakes that's important. 7) Work hard and keep going. 8) You got this
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How did you get 504 meters from 22/sin(2.5)? I plugged that into wolframalpha and it comes out to 36.8??
you probably have it in radians instead of degrees.
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wouldn't the force of friction in the last problem be positive since -1.89x10^5 *cos(180) becomes positive?
Hi, I just had question why to assign Kinetic energy = Potential energy?
Hi there! This answer is discussed at 1:31 in the video. Since the problem stated that the work done by friction is negligible, we can assume that there is not friction and that all of the kinetic energy will be converted into only potential energy and vice versa. Since these are the only types of energy in this problem, the potential energy has to be equal to the kinetic energy. Hopefully this helps!
12:20 how come a negative divided by a negative is equal to a negative? I get that common sense prevails but I'm still confused. Wait, someone else pointed this out below. Ok now I'm playing devil's advocate. Why WOULDN'T the force of friction be in the negative? I could've sworn in previous questions we use negative newtons for friction.
why cos180?
It's explained at around 11:03. Hopefully this helps!