Question

A car of mass 1000 kg moves on a rough road bel road friction is 10% of its weight and air friction is 2% of its way to keep the car of the uniform speed 36 km per hour the power required will be

Answer 1

force needed=10%(1000)+2%(1000)

F=100+20=120 N

P=Fv=120×36000/60×60

=120×10

=1200watt=1.2kW

Answer 2

Answer:

The power required is 12kwatt.

Explanation:

First, let's convert the velocity from km/h to m/s,

$V=36\times \frac{5}{18}=10m/s$        (1)

The weight of the car is,

$W=1000\times 10=10000N$    (2)

As per the question, road friction(fr) is 10% of its weight. So,

$f_r=10\% of Weight=\frac{10}{100}\times 10000=1000N$         (3)

And air friction(fa) is 2% of its weight,

$f_a=2\%of weight=\frac{2}{100}\times 10000=200N$           (4)

The net force (F) acting on the car is,

$F=f_r+f_a=1000+200=1200N$          (5)

The power(P) in terms of force and velocity is given as,

$P=F.V$              (6)

By using equations (1) and (5) in equation (6) we get;

$P=1200\times 10=12000watt=12kwatt$

Hence, the power required is 12kwatt.