Question

A 900 kg car is moving on a level ground. At a certain instant the car has a velocity of 6m/s and an acceleration of 2 m/s2. If the power delivered at that instant is 12,000W, what is the (Friction and drag) resistance force acting on the car?

Answer 1

The force of friction is 200 N

Explanation:

The instantaeous power of the car at the instant considered is

$P=12,000 W$

Since we are considering only one instant of time, we can consider the velocity of the car to be constant. Therefore, we can rewrite the power as:

$P=Fv$

where

F is the force applied by the engine

v = 6 m/s is the velocity of the car

Solving for F,

$F=\frac{P}{v}=\frac{12,000}{6}=2000 N$

The net force on the car instead is given by

$\sum F = ma=(900)(2)=1800 N$

where

m = 900 kg is the mass

$a=2 m/s^2$ is the acceleration

The net force is given by

$\sum F = F-F_f$

where

F = 2000 N is the force produced by the engine

$F_f$ is the force of friction

And solving for $F_f$,

$F_f=F-\sum F=2000-1800 = 200 N$

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