Answer:
(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m
(b) thermal energy was generated by friction is 1.88 x [tex]10^{5}[/tex] J
(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N
Explanation:
given information:
m = 750 kg
initial velocity, [tex]v_{0}[/tex] = 110 km/h = 110 x 1000/3600 = 30.6 m/s[tex]\frac{30.6^{2} }{2x9.8}[/tex]
initial height, [tex]h_{0}[/tex] = 22 m
slope, θ = 2.5°
(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?
according to conservation-energy
EP = EK
mgh = [tex]\frac{1}{2} mv_{0} ^{2}[/tex]
gh = [tex]\frac{1}{2} v_{0} ^{2}[/tex]
h = [tex]\frac{v_{0} ^{2} }{2g}[/tex]
= 47.6 m
(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?
thermal energy = mgΔh
= mg (h - [tex]h_{0}[/tex])
= 750 x 9.8 x (47.6 - 22)
= 188160 Joule
= 1.88 x [tex]10^{5}[/tex] J
(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?
f d = mgΔh
f = mgΔh / d,
where h = d sin θ, d = h/sinθ
therefore
f = (mgΔh) / (h/sinθ)
= 1.88 x [tex]10^{5}[/tex]/(22/sin 2.5°)
= 373 N
