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.
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.
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