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Rain water is collecting in an open carriage at the rate of 50 g/sec. the carriage is moving at a speed of 10 m/s .What is the extra force required to move

the carriage?



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Suppose there is very little kinetic  friction coefficient μ between the road and the carriage wheels.  So we neglect that.

Let the mass of carriage is M (at  time = 0)  with no water collected in it.  Let it move with a constant speed of 10 m/s at this instance.  The force required to keep it moving at constant speed is zero, if there is no rain.

Now the mass increases at 50 gram/sec = 0.050 kg/sec.  We neglect the impact of velocity of the rain in the vertical direction.  The rain drops have no velocity in the horizontal direction.

So in 1 sec, 0.050 kg rain water has to be given a velocity of  10 m/sec.  The momentum of  0.050 kg should change from 0 to 10 m/sec  in one second,  in horizontal direction.

    force needed = rate of change of momentum = 0.050 * 10 m/sec / 1 sec
                 = 0.5 Newtons.

If this much force is given (extra) then cart will move at constant speed under constant rain.  Otherwise, it will slowdown due to increase in mass.

1 5 1
please click on thank you link and select best answer
the ans is 5 N
as per our book but it may be wrong becoz max ans. of s.chand books are wrng
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