Question

. A truck of mass 5 ton is travelling on a
horizontal road with 36 km hrl, stops on
traveling 1 km after its engine fails suddenly.
What fraction of its weight is the frictional
force exerted by the road? If we assume that
the story repeats for a car of mass 1 ton i.e.,
car moving with same speed stops at similar
distance, how much will the fraction be?​

Answer 1

Answer:

Explanation:

The friction is \dfrac{1}{200}

Explanation:

Given that,

Mass of truck = 5 ton

Speed = 36 km/hr = 10 m/s

Distance = 1 km = 1000 m

We need to calculate the coefficient frictional

Using relation of frictional

\mu mg=ma

\mu=\dfrac{a}{g}

We need to calculate the fraction of its weight is the  frictional force exerted by the road

Using formula of fraction

\dfrac{F}{W}=\dfrac{\mu mg}{mg}

\dfrac{F}{W}=\mu

\dfrac{F}{W}=\dfrac{a}{g}

Put the value of a from equation of motion

\dfrac{F}{W}=\dfrac{\dfrac{v^2-u^2}{2s}}{g}

Put the value into the formula

\dfrac{F}{W}=\dfrac{100}{2\times10\times1000}

\dfrac{F}{W}=\dfrac{1}{200}

If we assume that the story repeats for a  car of mass 1 ton i.e., can moving with  same speed stops in similar distance same

The fraction of its weight and the  frictional force does not depend on the mass.

So, The fraction of its weight and the  frictional force for car is same.

Hence, The friction is 1/200