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Sand and conveyor belt

  • Thread starter PeteSeeger
  • Start date
1. Homework Statement
Sand runs from a hopper at constant rate dm/dt onto a horizontal
conveyor belt driven at constant speed V by a motor.
(a) Find the power needed to drive the belt.
(b) Compare the answer to (a) with the rate of change of kinetic
energy of the sand. Can you account for the difference?

2. Homework Equations
[tex]P=F{\cdot}v[/tex]
[tex]F=\dot{m}v+ma[/tex] (Since the sand's horizontal velocity component in our assumed reference frame is 0)
[tex]W=\Delta{T}=\int{Fdx}[/tex] (This is a definite integral but I don't know how to show bounds in latex)
3. The Attempt at a Solution
Alright, so part (a) is pretty straightforward. a is 0, just the mass is changing, so [tex]F=\dot{m}v[/tex] and then [tex]P=\dot{m}v^2[/tex]
So, now for part (b), which is the part that is confusing me, [tex]T=\frac{1}{2}mv^2[/tex] so [tex]\dot{T}=\frac{1}{2}\dot{m}v^2+mva[/tex] a is 0, so [tex]\dot{T}=\frac{1}{2}\dot{m}v^2[/tex]
So, now we find that the rate of change of kinetic energy is one half the power, which is what I presume they expected us to find by the wording of the question. But, when I start to think about an explanation, what I see is that the reason these aren't equal is because, mathematically, when we derive the equation [tex]W=\Delta{T}=\int{Fdx}=\frac{1}{2}mv_b^2-\frac{1}{2}mv_a^2[/tex] we assume that F=ma and m is a constant. But, if we rederive W with the correct assumptions for this problem [tex]W=\Delta{T}=\int{Fdx}=\int{(ma+\dot{m}v)dx}=\int{\dot{m}vdx}=mv_b^2-mv_a^2[/tex] we find that from our definition [tex]W=\Delta{T}[/tex] that the actual formula we should be using for this problem is [tex]T=mv^2[/tex] Now, if we use this equation for kinetic energy we derived from the correct assumptions for the problem, now we find that [tex]P=\dot{T}[/tex] So, I feel that the answer to part (b) is that the reason for the difference is that we are actually using the wrong equation for kinetic energy in this problem, and if we correct the formula, the two quantities are equal. Will someone tell me if all of this is correct, because I feel like I'm doing more than they've expected me to in this problem.
 

TSny

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So, I feel that the answer to part (b) is that the reason for the difference is that we are actually using the wrong equation for kinetic energy in this problem, and if we correct the formula, the two quantities are equal. Will someone tell me if all of this is correct, because I feel like I'm doing more than they've expected me to in this problem.
We don't want to change the definition of kinetic energy (##\frac{1}{2}mv^2##).

So, you have shown that the power input to the system is greater than the rate at which kinetic energy of the sand is increasing. Makes you wonder if there is some other form of energy in the system that is also increasing with time.
 
264
26
If an object of mass m moving with speed v collides with and sticks
to another stationary object of mass m then why can't you use
conservation of energy to find the final speed of the combination
of mass 2 m?
 

haruspex

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If an object of mass m moving with speed v collides with and sticks
to another stationary object of mass m then why can't you use
conservation of energy to find the final speed of the combination
of mass 2 m?
I'm not sure whether you ask that because you do not know or whether you are just asking PeteSeeger to answer it.
 
264
26
If an object of mass m moving with speed v collides with and sticks
to another stationary object of mass m then why can't you use
conservation of energy to find the final speed of the combination
of mass 2 m?
I'm not sure whether you ask that because you do not know or whether you are just asking PeteSeeger to answer it.
Pete Seeger seemed to imply that a solution should be obtainable using conservation of energy.
This is of course not true for inelastic collisions.
 

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