. 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?
Answers
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