Question

The engine of a 2000 kg Mercedes going up

Answer 1

Option (2) is correct.

Speed of car in upward direction = v x sin(theta)

Power of the wheel = m x g x v x sin (theta)

m= mass of the car

g = acceleration due to gravity

v = velocity of the car

v = Power/( m x g x sin(30))

= 100 x 1000 W / ( 2000 Kg x 10 m/s^2 x 0.5)

= 10m/s

= 36km/h

Answer 2

Answer:

2. 10 m/s

Explanation:

There are two forces acting on the car along the inclined plane:

- The driving force of the car, F, pulling upward along the ramp

- The component of the weight of the car, $mg sin \theta$, pulling downward along the ramp

In order to go at constant speed, the acceleration must be zero, so the net force must be zero. Therefore we can write:

$F-mgsin \theta =0$

From which we can find the driving force of the car:

$F=mgsin \theta=(2000 kg)(10 m/s^2)(sin 30^{\circ})=10,000 N$

The power of the engine is the product between force and speed of the car:

$P=Fv$

since we know the power, $P=100 kW=100,000 W$, we can find the speed:

$v=\frac{P}{F}=\frac{100,000 W}{10,000 N}=10 m/s$