(a) How high a hill can a car coast up (engine disengaged) if friction is neglig
ID: 1341909 • Letter: #
Question
(a) How high a hill can a car coast up (engine disengaged) if friction is negligible and its initial speed is 102 km/h?m
(b) If, in actuality, a 750 kg car with an initial speed of 102 km/h is observed to coast up a hill to a height 19.0 m above its starting point, how much thermal energy was generated by friction?
J
(c) What is the average force of friction if the hill has a slope 2.5° above the horizontal? (Explicitly show on paper how you follow the steps in the Problem-Solving Strategy for energy found on pages 159 and 160. Your instructor may ask you to turn in this work.)
N (down the slope) (a) How high a hill can a car coast up (engine disengaged) if friction is negligible and its initial speed is 102 km/h?
m
(b) If, in actuality, a 750 kg car with an initial speed of 102 km/h is observed to coast up a hill to a height 19.0 m above its starting point, how much thermal energy was generated by friction?
J
(c) What is the average force of friction if the hill has a slope 2.5° above the horizontal? (Explicitly show on paper how you follow the steps in the Problem-Solving Strategy for energy found on pages 159 and 160. Your instructor may ask you to turn in this work.)
N (down the slope)
Explanation / Answer
a) initial speed, v = 102 km/h
= 102*5/18
= 28.33 m/s
final potentail energy = initial kinetic energy
m*g*h = 0.5*m*v^2
h = 0.5*v^2/g
= 0.5*28.33^2/9.8
= 40.95 m
b) themral energy generated = workdone by friction
= m*g*(h - h')
= 750*9.8*(40.95 - 19)
= 161332.5 J
c) Workdone by friction = F*d
W_friction = F*(h - h')/sin(2.5)
F = W_friction*sin(2.5)/(h-h')
= 161332.5*sin(2.5)/(40.95 - 19)
= 320.6 N
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