A passenger is travelling a train moving at 40 ms-1.His suitcase is ke...
Answer :- b
Solution :- Retardation of train =20/4 = 5ms2
it act in the backward direction friction force on suitcase 5m newton , where m is the mass of suitcase in acts in the forward direction due to this force the suitcase has a tendency to slide forward if suitcase is not to slide then 5m = force f of friction
or 5m = mmg or m = 5/10
= 0.5
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A passenger is travelling a train moving at 40 ms-1.His suitcase is ke...
Solution:
Initial velocity of the train, u = 72 km/h = 20 m/s
Final velocity of the train, v = 0 km/h = 0 m/s
Acceleration of the train, a = (v-u)/t = (0-20)/5 = -4 m/s² (negative sign indicates retardation)
Let the coefficient of friction between the suitcase and the berth be μ.
The force of friction acting on the suitcase, f = μmg, where m is the mass of the suitcase and g is the acceleration due to gravity.
The force of friction opposes the motion of the suitcase and helps to prevent its sliding.
The maximum force of friction that can act on the suitcase without causing it to slide can be calculated as:
fmax = μmaxmg, where μmax is the maximum coefficient of friction.
To find the minimum coefficient of friction, we need to equate the force of friction with the maximum force of friction.
f = fmax
μmg = μmaxmg
μ = μmax
μmin = a/g
Substituting the values of a and g, we get:
μmin = -4/9.8 = -0.408
Since the coefficient of friction cannot be negative, the minimum coefficient of friction required to prevent the suitcase from sliding is 0.408.
Therefore, the minimum coefficient of friction between the suitcase and the berth should be 0.408 to prevent the suitcase from sliding during the retardation of the train.
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