Question

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An engine pulls a 1500 kg car on a level road at a constant speed of 5.0 m s against a frictional force of 500 N. What extra power has the engine develop in order to maintain the same speed of the car up an inclined plane having a gradient of 1 in 10 is

Answer 1

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Power = Force × Velocity

= 500 N × 5 m/s

= 2500 Watt

= 2.5 kilowatt

Power expended by engine is 2.5 kW

Answer 2

1] Power developed on the plane surface is

P = Fv

= 500*5 = 2.5 kW

2] So now power developed on inclined surface will be.....

A) Total force acting on the body according to the diagram is Mgsinθ + Frictional force (Fk)

B) Gradient of 1 in 10 means at every 10 units (in x direction) there is rise in 1 unit (in y directions). So if we resolve this by tanθ=1/10

θ=5.71°

So, Mgsin(5.71) = Mg(0.0995)

= 1500*9.8*0.0995 = 1462.65 N

So total force will be 1462.65+500

= 1962.65 N

Condition is given to keep the speed same i.e. 5m/s

Therefore, Power = 1962.65*5

= 9813.25 Watt

So Extra power is 9813.25-2500

= 7313.25 Watt = 7.3kW

Hope it helps....:)