# Physics

(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 2.5°
above the horizontal?

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1. 110 km/hr = 110,000 m/3600 s = 30.6 m/s

m g h = (1/2) m v^2

9.81 h = .5 (30.6)^2

h = 47.7 meters

(b) loss of energy = m g (47.7-22)
= 750(9.81)(25.7) = 189,000 Joules lost

(c) work done = force in direction of motion * distance moved

189,000 = F (22/sin2.5)

F = 375 N

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posted by Damon
2. Thank You

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3. You are welcome.

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posted by Damon
4. I disagree with the answer to (c). The distance traveled (d) is not the same as the height the car has reached at the end of its coasting. d is actually the hypotenuse. so h=dsin(2.5) thus d=h/sin(2.5), so d=22/0.424 which is 52.1, the distance traveled. Plugging this into the work done by friction,(Wf), Wf= Fdcos(2.5), F= Wf / dcos(2.5),F = 189,000/52.06cos2.5 = 3634N.

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posted by Steve
5. Hi, your response to question c) is right but for the wrong reasons and could be easily misconstrued as the correct method. Taking the value of energy loss from the previous question as the work done by friction we can use the equation for work with respect to the force, displacement and angle at which the force is applied. The displacement is in fact the hypotenuse, as is given by (22/sin(2.5 degrees)). The primary mistake however is ignoring the fact that the force of friction is acting in the opposite direction of the car, and therefore the angle that should be used in the work equation is 180 degrees, not the 2.5 as some others have suggested. This gives the value of F to be -375N, which when looking solely for the magnitude gives the force of friction to be 375N.

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posted by Kyle

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