Question

An engine pulls a car of mass 1500kg on a level road at a constant speed of 5m/s. The Extra power must the engine develop to maintain the same speed up an inclined plane having a gradient of 1 in 10 is (g = 10m/s²)​

Answer 1

Explanation:

Q) Power =F× velouty mo and =0. F=500 N = frictional =500×5 = gradient in 10

=2500W=2.5kW0.10

in an thanne F=50% toru foravity (F1)

Answer 2

Topic :-

Mechanics

Given :-

An engine pulls a car of mass 1500kg on a level road at a constant speed of 5m/s.

Gradient of Inclined Plane is 1 in 10.

To Find :-

The extra power must the engine develop to maintain the same speed up an inclined plane.

Concept Used :-

Power = Force × Velocity

Work done by a mass moving with constant speed is zero.

Force required to push object up on an incline is $mgsin\theta$

Solution :-

It is given that gradient of inclined plane is 1 in 10.

$It\:means\:tan\theta=\dfrac{1}{10}$

$tan\theta=sin\theta ,\:for\:small\:value\:of\:tan\theta$

Mass of object, m = 1500 kg

g = 10 m/s²

Velocity of object, v = 5 m/s

Power exerted by engine to pull a car of mass 1500kg on a level road at a constant speed of 5m/s is Zero as net acceleration is zero.

Power exerted by the engine to maintain the same speed up an inclined plane.

$Power = Force \times Velocity$

$Power=mgsin\theta \times Velocity$

Put values,

$Power=1500 \times 10 \times \dfrac{1}{10} \times 5$

$Power=1500 \times 10 \times \dfrac{1}{10} \times 5$

$Power=7500\:Watts$

$Power=7.5\:kW$

Calculating Extra Power developed by Engine,

Extra Power = 7.5 kW - 0 kW

Extra Power = 7.5 kW

Answer :-

The extra power must the engine develop to maintain the same speed up an inclined plane is 7.5 kW.