Question

. (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?

Answer 1

(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
`10^(5)` J

(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N

Step-by-step explanation:

given information:

m = 750 kg

initial velocity,
`v_(0)` = 110 km/h = 110 x 1000/3600 = 30.6 m/s
`(30.6^(2) )/(2x9.8)`

initial height,
`h_(0)` = 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 =
`(1)/(2) mv_(0) ^(2)`

gh =
`(1)/(2) v_(0) ^(2)`

h =
`(v_(0) ^(2) )/(2g)`

= 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 -
`h_(0)`)

= 750 x 9.8 x (47.6 - 22)

= 188160 Joule

= 1.88 x
`10^(5)` 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
`10^(5)`/(22/sin 2.5°)

= 373 N