Question

A bus wulghing 100 quintals moves on a rough road with a constant speed of 72 km PH The friction of the road is 9% of its weight and that of air is 1% of its
weight. Then the power of the engine is​

Answer 1

I hope its helpfull to u but it is blur sory

Answer 2

Concept:

• Calculating net forces
• Air resistance and frictional force

Given:

• mass of the bus = 100 quintal = 100*100 kg = 10000 kg
• Speed of the bus = 72 km/h = 20m/s
• Friction of road = 9% = 0.09 of the weight of the bus
• Air resistance = 1% = 0.01 of the weight of the bus

Find:

• The power of the engine

Solution:

Weight of the bus = mg = 10000 *10 = 10^5 N

Frictional force = 0.09*mg = 9000 N

Air resistance = 0.01 mg = 1000 N

Net opposing force = frictional force + air resistance = 9000 + 1000 = 10000N

Power = Force * speed

P = Fv

P = 10000 * 20 = 2*10^5 W = 200kW

The power of the engine is 200kW.

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