A 2800-lbm car climbs a 1160 ft. long uphill road which is inclined at 15o (to the horizontal) and covers the distance in 12 sec. Determine the power required if the car covers the distance (a) at constant velocity, (b) from an initial velocity, at the bottom of the hill, of 10 mph to a final velocity of 50 mph at the top of the hill and (c) from an initial velocity of 45 mph to a final velocity of 15 mph. Neglect the effects of friction and wind resistance.

a) P = 70054.3 W,  b)  P = 18820 W,  c)   P = 14116.7 W

Explanation:

Power is defined as work per unit of time

         P = W / t = F x / t

         P = F v

a) in this case the velocity is constant, let's use the equilibrium relation to find the force.

Let's set a reference system with the x axis parallel to the plane

        F - Wₓ = 0

         F = Wₓ

with trigonometry let's decompose the weight

        sin θ = Wₓ / W

        Wₓ = W sin θ

          F = W sin 15

          F = 2800 sin 15

          F = 724.7 lb

we look for the speed, as it rises with constant speed we can use the relations of uniform motion

          v = x / t

          v = 1160/12

          v = 96.67 ft / s

we calculate the power

          P = 724.7 96.67

          P = 70054.3 W

b) In this case, the speed of the vehicle changes during the ascent, so we use the relationship between work and the change in kinetic energy

           W = ΔK

           W = ½ m v_f² - ½ m v₀²

   let's reduce to the SI system

           v₀ = 10 mph (5280 ft / 1 mile) (1h / 3600 s = 14.67 ft / s

           v_f = 50 mph (5280 ft / 1 mile) (1 h / 3600s) = 73.33 ft.s

 mass :         m = w / g

           W = ½ 2800/32 (73.33² - 14.67²)

           W = 225841 J

we calculate the average power

           P = W / t

           P = 225841/12

           P = 18820 W

c) we repeat the previous procedure

          v₀ = 45 mph = 66 ft / s

          v_f = 15 mph = 22 ft / s

          W = ½ 2800/32 (22² - 66²)

          W = -169400 J

          P = W / t

          P = 169400/12

          P = 14116.7 W

hotter stars = bluer stars

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)