Question

A 1830-kg car experiences a combined force of air resistance and friction that has the same magnitude whether the car goes up or down a hill at 34.5 m/s. Going up a hill, the car's engine needs to produce 74.7 hp more power to sustain the constant velocity than it does going down the same hill. At what angle is the hill inclined above the horizontal?

Answer 1

let the angle of incline is ""

m = mass of the car = 1830 kg

consider the situation while going Up :

F = applied force

f = combined force by friction and air resistance

a = acceleration of car = 0 (since car moves at constant velocity)

v = velocity of car = 34.5 m/s

P = power used while going up

parallel to incline , force equation is given as

F - f - mg Sin = ma

F = f + mg Sin eq-1

Power used going Up is given as

P = F v

using eq-1

P = ( f + mg Sin) v eq-2

consider the situation while going down :

F' = applied force

f = combined force by friction and air resistance

a = acceleration of car = 0 (since car moves at constant velocity)

v = velocity of car = 34.5 m/s

P' = power used while going down

parallel to incline , force equation is given as

F' - f + mg Sin = ma

F' = f - mg Sin eq-3

Power used going Up is given as

P' = F' v

using eq-1

P' = ( f - mg Sin) v eq-4

Given that :

P - P' = 74.7 hp

P - P' = 74.7 x 746 Watt since 1 hp = 746 Watt

using eq-2 and eq-4

( f + mg Sin) v - ( f - mg Sin) v = 74.7 x 746

2 mg Sin v = 74.7 x 746

2 (1830 x 9.8) Sin (34.5) = 74.7 x 746

Sin = 0.045

= Sin^-1(0.045)

= 2.6 deg