Question

A car has an initial speed of 118 km/h and climbs up an incline with its engine DISENGAGED (no engine force). DON"T use scientific notation. Angle is 33 degrees.

(a) If work done by friction is negligible, How high (the h in the figure) a hill can the car coast up (engine disengaged) before coming to a stop?

(b) If, in actuality, a 700-kg car with an initial speed of 118 km/h is observed to coast up a hill to a vertical height hf of only 44.8 m above its starting point before coming to a stop, how much thermal energy was generated by friction?

(c) What is the magnitude of the average force (in Newtons) of kinetic friction fk if the hill has a slope 33 degrees above the horizontal?

(d) What is the magnitude of the normal force (in Newtons)? What is the coeficient of kinetic friction?

Answer 1

a)

v_o = initial speed = 118 km/h = 32.78 m/s

h = height gained

m = mass of the car

Using conservation of energy

Kinetic energy of the car = Potential energy at the top

(0.5) m v_o² = mgh

h = (0.5) v_o²/g

h = (0.5) (32.78)²/(9.8)

h = 54.82 m

b)

h_a = actual height gained = 44.8 m

W_t = thermal energy generated

Using conservation of energy

Kinetic energy at the bottom = potential energy at he top + thermal energy generated

(0.5) m v² = mgh + W_t

(0.5) (700) (32.78)² = (700 x 9.8 x 44.8) + W_t

W_t = 6.88 x 10⁴ J

c)

d = length of the incline

Sin33 = h/d

Sin33 = 44.8/d

d = 82.3 m

f = magnitude of average force of kinetic friction

Work done by the frictional force is given as

W_t = fd

6.88 x 10⁴ = f (82.3)

f = 836 N

d)

N = normal force

Normal force is given as

N = mg Cos33

N = (700 x 9.8) Cos33

N = 5753.3 N

Coefficient of kinetic friction is given as

= f/N

= 836/5753.3

= 0.15