Question

a) If 8.00 hp are required to drive a 1800-kg automobile at 54.0 km/h on a level road, what is the total retarding force due to friction, air resistance, and so on?

b) What power is necessary to drive the car at 54.0 km/h up a 10.0% grade (a hill rising 10.0 m vertically in 100.0 m horizontally)?

c) What power is necessary to drive the car at 54.0 km/h down a 1.00 % grade?

d) Down what percent grade would the car coast at 54.0 km/h ?

Answer 1

Part A.

Power is given by:

Power = Force*Velocity = F*V

F = P/V

Given that

P = 8.00 hp = (8.00 hp)*(746 W/1 hp) = 5968 W

V = 54.0 km/hr = (54.0 km/hr)*(1 hr/3600 sec)*(1000 m/1 km) = 15 m/sec

So,

F = 5968/15 = 397.87 N = 398 N

Retarding Force = F = 398 N

Part B.

Given that 10.0% grade road, So Now Net Power will be:

P_net = P + P1

P = Power required on a level road = 8.00 hp

P1 = Power required on a 10.0% grade road = Power required against gravity

P1 = W*V1

W = Weight of car = m*g

V1 = speed of car for upward motion on the incline = V*sin A

Given that road has 10.0% grade, So

A = arctan (Vertical height/horizontal distance) = arctan (10/100) = 5.71 deg

So,

P1 = m*g*V*sin A

P1 = 1800*9.81*15*sin 5.71 deg = 26352.8217 W = (26352.8217 W)*(1 hp/746 W)

P1 = 35.34 hp

So,

P_net = 8.00 hp + 35.34 hp = 43.34 hp

Part C.

Given that 1.00% grade road and car is traveling downward, So Now Net Power will be:

P_net = P - P1

P = Power required on a level road = 8.00 hp

P1 = Power required on a 1.00% grade road = Power required against gravity

P1 = W*V1

W = Weight of car = m*g

V1 = speed of car for downward motion on the incline = V*sin A

Given that road has 1.00% grade, So

A = arctan (Vertical height/horizontal distance) = arctan (1/100) = 0.573 deg

So,

P1 = m*g*V*sin A

P1 = 1800*9.81*15*sin 0.573 deg = 2648.85095 W = (2648.85095 W)*(1 hp/746 W)

P1 = 3.55 hp

So,

P_net = 8.00 hp - 3.55 hp = 4.45 hp

Part D.

When the car is travelling downward, then for car to coast at 54.0 km/hr, when all the work-done is done by gravity

Power = 8.00 hp

P = m*g*V*sin A1

sin A1 = P/(m*g*V)

A1 = arcsin (P/(m*g*V))

Using known values:

A1 = arcsin (8.00*746/(1800*9.81*15))

A1 = 1.291 deg

Now grade of road will be:

grade = tan A1 = tan 1.291 deg = 0.0225

grade = 0.0225 = 2.25%

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