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Question
Solve the following problem.
A truck of mass 5 ton is travelling on a horizontal road with 36 km hr-1, 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 at the same speed stops at a similar distance, how much will the fraction be?
Solution
Given:
mtruck = 5 ton = 5000 kg,
mcar = 1 ton = 1000 kg,
u = 36 km/hr = 10 m/s,
v = 0 m/s, s = 1 km = 1000m
To find: Ratio of the force of friction to the weight of the vehicle
Formulae:
i. v2 = u2 + 2as
ii. F = ma
Calculation:
From formula (i),
2 × atruck × s = v2 - u2
∴ 2 × atruck × 1000 = 02 - 102
∴ 2000atruck = - 100
∴ atruck = - 0.05 m/s2
Negative sign indicates that velocity is decreasing
From formula (ii),
Ftruck = mtruck × atruck
= 5000 × 0.05
= 250 N
∴ `"F"_"truck"/"weight"_"truck" = 250/(5000 xx 10) = 1/200`
From formula (i),
2 × acar × s = v2 - u2
∴ 2 × acar × 1000 = 02 - 102
∴ 2000 acar = - 100
∴ acar = - 0.05 m/s2
From formula (ii),
F = mcar acar = 1000 × 0.05 = 50 N
∴ `"F"_"car"/"weight"_"car" = 50/(1000 xx 10) = 1/200`
The frictional force acting on both the truck and the car is `1/200` of their weight.
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