gravel is dropped on a conveyor belt at the rate of 2 kg/s .the extra force required to keep the belt moving at 3m/s. pls explain clearly
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Answer:
Refer to the figure. If there were no friction, it would make no difference how the rise occurred, only the amount of rise. In that case vhole=√(vputt2-2gh). However, there is friction which will slow the ball further and the frictional force on a slope (f=μmgcosθ) will be different than on level ground (f=μmg); here μ is the coefficient of friction (essentially, I believe, what is measured by the "stimp meter"), m is the mass of the ball, and g is the acceleration due to gravity. Since the frictional force is smaller on the slope, you might think that less energy is lost to friction on the slope. But, what matters is not the force but its product with the distance s over which it acts, specifically the work done by friction is Wf=-fs where the negative sign indicates that the friction takes energy away from the ball. When the ball is moving forward along along the segment L2, the distance it travels is s=L2/cosθ and so the work done is Wf=fs=(μmgcosθ)(L2/cosθ)=μmgL2, exactly the same as if the slope was not there. The total work done by friction, regardless of the path, is Wf=μmgL. The final velocity is then vhole=√[vputt2-2g(h+μL)].