A barge loaded with scrap iron is floating in a lock. If all the scrap iron falls off the barge into the water in the lock, will the water level of the lock go up, down, or remain the same?
Consider a person pulling a sled at a constant velocity of 0.5 meters/sec. while snow is falling and accumulating on the sled at the rate of 0.1 Kg per second. If we ignore the drag of the sled runners on the snow on the ground, what is the horizontal component F of the force the person must exert to keep the sled moving?
Should we use Momentum or Energy?
Conservation of Energy Method. Every second the Kinetic energy of the sled/snow is increased by 1/2mv2 = 1/2*0.1*0.52 = 0.0125 kg m2/s3 or joules/sec or watts. This should be equal to the work done by the force F per second which is Fv = F*0.5 kg m2/s3 or watts. Thus F = 0.0125/0.5 N = 0.025 N.
Momentum Method. Every second the momentum of the sled/snow is increased by m v or 0.1*0.5 = 0.05 kg m/s2. This should be equal to the impulse of the force F per second which is F*1 second = F kg m/s2. Thus F = 0.05 N.
Whoops this is exactly twice what we found by the energy method!
Both answers can't be right, so is either correct, and if so, what's wrong with the other method?